Solution for Introduction to Environment Engineering and Science 2nd edition by Gilbert M. Masters
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About This Presentation
Complete Solution for Introduction to Environment Engineering and Science 2nd edition by Gilbert M. Masters
Slide Content
SOLUTIONS MANUAL
INTRODUCTION TO
ENVIRONMENTAL
ENGINEERING
and SCIENCE
SECOND EDITION
GILBERT M. MASTERS
INTRODUCTION TO
ENVIRONMENTAL
ENGINEERING
and SCIENCE
SECOND EDITION
GILBERT M. MASTERS
SOLUTIONS MANUAL ‘
INTRODUCTION TO
ENVIRONMENTAL
ENGINEERING
and SCIENCE
SECOND EDITION
GILBERT M. MASTERS
Dep of Gil and Environmental Engncrn,Senford Univers
E ranrice nats, ur Se Rn 01856
INTRODUCTION TO
ENVIRONMENTAL
ENGINEERING
and SCIENCE
SECOND EDITION
GILBERT M. MASTERS
Dep of Gil and Environmental Engncrn,Senford Univers
E ranrice nats, ur Se Rn 01856
Executive Editor: Bill Siequir
sito Asunto Meg Weit
Production Eaton Shee Ockey
Special Projets Manager Barbara A. Murray
Stpplement Cover Manager Pel Gourkan
Manfacuing Boyer: Jule Meche
©1986 by Prenton Mal, Ine
Simon & Schuster! A Viacom Company
Upper Sade River, N 07458
Aitighs reserved. No pat of his book may be
‘proce, lay ores or y any means,
bout permisos l ring rom the publisher.
Printed inthe Usted States of America
987654321
ISBN 0-13-889064-1
rente Hal Imemationa! (UK) Lite, London
rente Hal ol Aula Py. Limite, Sydney
Premio Hall Canada, In, Toronto
Pre Hal apaccamescana, SA, Mexico
Bret Half Inds Private Limited, New De
Pret Hall of apa, Ic, Tokyo
Simon & Schuster Aas ie Lid, Singapore
ElioraPrenue-Haldo Bras ida, Ri de Jeneto
sito Asunto Meg Weit
Production Eaton Shee Ockey
Special Projets Manager Barbara A. Murray
Stpplement Cover Manager Pel Gourkan
Manfacuing Boyer: Jule Meche
©1986 by Prenton Mal, Ine
Simon & Schuster! A Viacom Company
Upper Sade River, N 07458
Aitighs reserved. No pat of his book may be
‘proce, lay ores or y any means,
bout permisos l ring rom the publisher.
Printed inthe Usted States of America
987654321
ISBN 0-13-889064-1
rente Hal Imemationa! (UK) Lite, London
rente Hal ol Aula Py. Limite, Sydney
Premio Hall Canada, In, Toronto
Pre Hal apaccamescana, SA, Mexico
Bret Half Inds Private Limited, New De
Pret Hall of apa, Ic, Tokyo
Simon & Schuster Aas ie Lid, Singapore
ElioraPrenue-Haldo Bras ida, Ri de Jeneto
SOLUTIONS FOR CHAPTER 1
SOLUTIONS FOR CHAPTER 2
SOLUTIONS FOR CHAPTER 3
SOLUTIONS FOR CHAPTER 4
SOLUTIONS FOR CHAPTER 5
SOLUTIONS FOR CHAPTER 6
SOLUTIONS FOR CHAPTER 7
SOLUTIONS FOR CHAPTER 8
SOLUTIONS FOR CHAPTER 9
CONTENTS
11
21
31
aa
51
64
74
aa
91
SOLUTIONS FOR CHAPTER 2
SOLUTIONS FOR CHAPTER 3
SOLUTIONS FOR CHAPTER 4
SOLUTIONS FOR CHAPTER 5
SOLUTIONS FOR CHAPTER 6
SOLUTIONS FOR CHAPTER 7
SOLUTIONS FOR CHAPTER 8
SOLUTIONS FOR CHAPTER 9
CONTENTS
11
21
31
aa
51
64
74
aa
91
SOLUTIONS FOR CHAPTER 1
ppm x mol wt
24414
23.15 Plaim)
TK) * Tam
à mat (25°C, Lamy x DBA, AS _
18/m (at 250C, | atm) > (273.15+25)
1.1 Ozone 20.08 ppm; mg/m? = as
157 mg? = 157 gm
b. In Denver, at 15°C and 0.82 atm:
3 008 x(3x 16)
273.15 _ 0.82
BET za
x 20.133 mg/m’ = fm?
(273.15+15) 1 um 123 ah
1.2 Exhaustat 1% CO,259C, 1 atm 1% CO = LB CO pam CO 516" pom
(00 paris air ” 10°pansair
„E10 x(12+16) 285
en 2a * (273.15+25)
= 11,445 mg/m?
13 4001gmÓ of $02 at 25°C, 1 atm:
24465 x $0,(mg/m") _ 24.465 x 0.400
pris Im’) 28465 70400 6 15 09m
mol we (6242516)
YES, exceeds the air quality standard of 0.14 ppm.
1.4 Motoreyele emiting 20 g/mi of CO:
ma 1m
a v= x 5m x 12 x ol x LEE
208 CO Fay SL Imola dofr
e oppmco: 22260. „009 m? CO FA.
10° mw air Vm’air
m polluted 9666 m?
mile mix 1609 mimi
212m /m
Page 1.1
ppm x mol wt
24414
23.15 Plaim)
TK) * Tam
à mat (25°C, Lamy x DBA, AS _
18/m (at 250C, | atm) > (273.15+25)
1.1 Ozone 20.08 ppm; mg/m? = as
157 mg? = 157 gm
b. In Denver, at 15°C and 0.82 atm:
3 008 x(3x 16)
273.15 _ 0.82
BET za
x 20.133 mg/m’ = fm?
(273.15+15) 1 um 123 ah
1.2 Exhaustat 1% CO,259C, 1 atm 1% CO = LB CO pam CO 516" pom
(00 paris air ” 10°pansair
„E10 x(12+16) 285
en 2a * (273.15+25)
= 11,445 mg/m?
13 4001gmÓ of $02 at 25°C, 1 atm:
24465 x $0,(mg/m") _ 24.465 x 0.400
pris Im’) 28465 70400 6 15 09m
mol we (6242516)
YES, exceeds the air quality standard of 0.14 ppm.
1.4 Motoreyele emiting 20 g/mi of CO:
ma 1m
a v= x 5m x 12 x ol x LEE
208 CO Fay SL Imola dofr
e oppmco: 22260. „009 m? CO FA.
10° mw air Vm’air
m polluted 9666 m?
mile mix 1609 mimi
212m /m
Page 1.1
1.5 Air density with 79% No and 21% O2:
Nop 29m N, 288, Liat Ns
Tar mo * 22414x107m Ny,
o: VIN, ,328 , Lol
#5 Tair mot * 224100 mW,
Total = 987 + 300 = 1287 g/m? = 1.287 kg/m?
1.6 Mixing 10 MGD, 3.0 mg/L, with 5 MGD, 10.0 mg/L
a. 10MGD x3.0 mg/L + SMGD x 10.0 mg/L = (10+5)MGDxC mg/L
C=80/15 = 533 my
533 DE al Like ob
o. 533 PE x 15x 10° E x 3.785 — x TE x 225 = 666 1b/de)
L day qa Img * kg z
17
500 ppm
25 m3 /s, 400 ppm — A RCE 0 pen
5 m3/s, 2000 ppm À
25m%/s x400mpL + Sls x 2000 ML _ 7 y
(25 + Sms eel
Upstream of take-out: C=
500 mg/L x(Q + FQ)m'/s= 667 mg/L xQm*/s
S00(1+ F)= 667
LE 120599 (ti 15 pew
Page 1.2
Nop 29m N, 288, Liat Ns
Tar mo * 22414x107m Ny,
o: VIN, ,328 , Lol
#5 Tair mot * 224100 mW,
Total = 987 + 300 = 1287 g/m? = 1.287 kg/m?
1.6 Mixing 10 MGD, 3.0 mg/L, with 5 MGD, 10.0 mg/L
a. 10MGD x3.0 mg/L + SMGD x 10.0 mg/L = (10+5)MGDxC mg/L
C=80/15 = 533 my
533 DE al Like ob
o. 533 PE x 15x 10° E x 3.785 — x TE x 225 = 666 1b/de)
L day qa Img * kg z
17
500 ppm
25 m3 /s, 400 ppm — A RCE 0 pen
5 m3/s, 2000 ppm À
25m%/s x400mpL + Sls x 2000 ML _ 7 y
(25 + Sms eel
Upstream of take-out: C=
500 mg/L x(Q + FQ)m'/s= 667 mg/L xQm*/s
S00(1+ F)= 667
LE 120599 (ti 15 pew
Page 1.2
1.8
sont (TO seiner
À 5 m8 /s, 100 mg/L, K=0.25/d
Input = Output + Decay , where decay
cv
0.25/day
RC (git) Om
ZA held x 3600 Thr
Sis x 100 mg/L = 55 mx C (mg +
500= 550+289¢ 596 mg/L
19
1160 ae oer
20 mg/L sus a >
Ka 03078
Lake I: Input = Output + KCV
1 MGD x 20 mg/L = 1 MGD x C1 + 031day x SMOxCI
2
ee «80m
Lake 2: 1 MOD x 80 mg/L = 1 MGDx C2 + 03443 MG x C2
D 42 mL
1.10
01 m/s
30 mg/L —
7]
Input = Output + KCV
01 m°sx30 ml = 0.1 mls x 10 mgL + =>) lomgLx Vm"
24 held x 3600 sh,
30-10 _ 6 400m
231x10"
Page 13
sont (TO seiner
À 5 m8 /s, 100 mg/L, K=0.25/d
Input = Output + Decay , where decay
cv
0.25/day
RC (git) Om
ZA held x 3600 Thr
Sis x 100 mg/L = 55 mx C (mg +
500= 550+289¢ 596 mg/L
19
1160 ae oer
20 mg/L sus a >
Ka 03078
Lake I: Input = Output + KCV
1 MGD x 20 mg/L = 1 MGD x C1 + 031day x SMOxCI
2
ee «80m
Lake 2: 1 MOD x 80 mg/L = 1 MGDx C2 + 03443 MG x C2
D 42 mL
1.10
01 m/s
30 mg/L —
7]
Input = Output + KCV
01 m°sx30 ml = 0.1 mls x 10 mgL + =>) lomgLx Vm"
24 held x 3600 sh,
30-10 _ 6 400m
231x10"
Page 13
1.
ams
som
zo
Core
CO Input rate = Output rate
60 ke/s x 105 mg/kg = 20 x 108 m x 250 m x 2 m/s x (CO) mg?
60 x 10° >
N o mim
© 20,000 x 250 x 2 su
1.12
100 kan
e. Soo] E
OS um
oi.
Bie
Inputrate = Output rate + KCV
Input rate = 10 kg/s x 109 pg/kg = 10 x 109 ug/s
Output rate = 100 x 103 m x 103 m x 4 mls x (um) =4x 105C pels
2e
ecaÿ rate = ———— gm?) x 105m x 105 m x 103 m = 5.55 x 108C pels
Decay ate = ¿LC (al) 105 mx 105 555x108 Cu
10 x 10” =
TEEN
al (120)
113
ra
$ = 10x 109 pgs
= 1misx 105 mx 103 m= 1 x 108 ms (new value at 1 m/s wind)
ZUNE 195m x 105 mx 108 m= 5.55 x 108 m/s
KY = 3600 sr
ATP
TA 1526 pat
Cl) =[C, -C Jet +Q/ va] + Ca (129)
caos 1525 E a] saan?
= 123 ugmd
Page 1.4
ams
som
zo
Core
CO Input rate = Output rate
60 ke/s x 105 mg/kg = 20 x 108 m x 250 m x 2 m/s x (CO) mg?
60 x 10° >
N o mim
© 20,000 x 250 x 2 su
1.12
100 kan
e. Soo] E
OS um
oi.
Bie
Inputrate = Output rate + KCV
Input rate = 10 kg/s x 109 pg/kg = 10 x 109 ug/s
Output rate = 100 x 103 m x 103 m x 4 mls x (um) =4x 105C pels
2e
ecaÿ rate = ———— gm?) x 105m x 105 m x 103 m = 5.55 x 108C pels
Decay ate = ¿LC (al) 105 mx 105 555x108 Cu
10 x 10” =
TEEN
al (120)
113
ra
$ = 10x 109 pgs
= 1misx 105 mx 103 m= 1 x 108 ms (new value at 1 m/s wind)
ZUNE 195m x 105 mx 108 m= 5.55 x 108 m/s
KY = 3600 sr
ATP
TA 1526 pat
Cl) =[C, -C Jet +Q/ va] + Ca (129)
caos 1525 E a] saan?
= 123 ugmd
Page 1.4
1.14
100m3/d x 1Omg/L = 100m3dxC C= 10 mg/L
b. Change input concentration suddenly to 100 mg/L, Ces = 100 mg/L, C (7days) =
(C, -C.Jesp[=(K + Q VI] + Ca (29)
(om
er 1200 m’
100 m3/d x 10 mg/L = 100 m3/d x C mg/L. + 0.20/ x C mg/L x 1200 m3
Das age
1.15
now K=020/day
100 x 10
100 + 020 x 1200
94 mg.
. Change input suddenty to 100 mg/L, C(7 days) = ?
s 100 md x 100 mg/L.
Q+KV 7 100m 18 + 0.2018 x 1200m
29.4 mglL
Cle) =[C, —C. Jexpl-(K + VI] + Co
100 m°/d
tet a] + wenn
CA) «poor (E
= 258mgL
Page 15
100m3/d x 1Omg/L = 100m3dxC C= 10 mg/L
b. Change input concentration suddenly to 100 mg/L, Ces = 100 mg/L, C (7days) =
(C, -C.Jesp[=(K + Q VI] + Ca (29)
(om
er 1200 m’
100 m3/d x 10 mg/L = 100 m3/d x C mg/L. + 0.20/ x C mg/L x 1200 m3
Das age
1.15
now K=020/day
100 x 10
100 + 020 x 1200
94 mg.
. Change input suddenty to 100 mg/L, C(7 days) = ?
s 100 md x 100 mg/L.
Q+KV 7 100m 18 + 0.2018 x 1200m
29.4 mglL
Cle) =[C, —C. Jexpl-(K + VI] + Co
100 m°/d
tet a] + wenn
CA) «poor (E
= 258mgL
Page 15
1.16
300 Btu/r2-
Sel gem
~ Tat
46
Nam Ti ne son
Rate of energy absorbed = Rate of change of stored energy
co x300 te nanaan = 10 Basse canti
Pr
ATT 834% 60 oor
117
week evaporation
Bu
21000 x 624 By x 10505 = 81.9 x 10* Em season
sed
xx
Bu. $1000
oversaves: 2 x819%
S 3 10 Jason O Bus
= $5461 yr
‘YES a $500 cover does pay for itself in less than one season.
Page 16
300 Btu/r2-
Sel gem
~ Tat
46
Nam Ti ne son
Rate of energy absorbed = Rate of change of stored energy
co x300 te nanaan = 10 Basse canti
Pr
ATT 834% 60 oor
117
week evaporation
Bu
21000 x 624 By x 10505 = 81.9 x 10* Em season
sed
xx
Bu. $1000
oversaves: 2 x819%
S 3 10 Jason O Bus
= $5461 yr
‘YES a $500 cover does pay for itself in less than one season.
Page 16
118
NS
ns
a RN
a. Rate of heat added to cooling water = rate of change of stored energy = m € AT
TE a cre
w
= 200x 10°
2184: 10
17.8 x 10 ka/s x 10’mkg = 47.8 mls
nuclea plank cooling water _ 47.8m'/
—tuclear plant cooling water _ _ EMS 1 177 (= 18% moro)
Zeal plant cooling water (Ex 0) "406 mors 17 18% more)
b. River temperature rises by AT asit receives 2000 MW of heat:
cooling waterheat gain rate _ 2000 x 108 is
me TOO Tex A184 JC x 10 Kam?
=48'C
1.19 Moisture condensing releases latent heat = 5 mL x 1 /mLx23KI/3=125KJ
¿AT 20354 Kg x 418K OC x AT = 12510
125
AT aaa aa = 84°C
1.20 Energy needed to vaporize | kg of water at 15°C (Table 1.4)= 2465 id
To raise 1 kg 3000 m requires:
98N U
x LL 2 29,4005 = 29480
kg N-m y
3000 mx 1 kg x
Energy tovaporize water _ 2465 KI
Energy toraiseit3km 29.4 K
= 8:1
Page 17
NS
ns
a RN
a. Rate of heat added to cooling water = rate of change of stored energy = m € AT
TE a cre
w
= 200x 10°
2184: 10
17.8 x 10 ka/s x 10’mkg = 47.8 mls
nuclea plank cooling water _ 47.8m'/
—tuclear plant cooling water _ _ EMS 1 177 (= 18% moro)
Zeal plant cooling water (Ex 0) "406 mors 17 18% more)
b. River temperature rises by AT asit receives 2000 MW of heat:
cooling waterheat gain rate _ 2000 x 108 is
me TOO Tex A184 JC x 10 Kam?
=48'C
1.19 Moisture condensing releases latent heat = 5 mL x 1 /mLx23KI/3=125KJ
¿AT 20354 Kg x 418K OC x AT = 12510
125
AT aaa aa = 84°C
1.20 Energy needed to vaporize | kg of water at 15°C (Table 1.4)= 2465 id
To raise 1 kg 3000 m requires:
98N U
x LL 2 29,4005 = 29480
kg N-m y
3000 mx 1 kg x
Energy tovaporize water _ 2465 KI
Energy toraiseit3km 29.4 K
= 8:1
Page 17
1.21 Condensation of 1 tb of water releases 1060 Btu (Table 1.4)
tal energy of Ki 643 Bro
Potential energy of 1 b at 5000 ft = 5000 fi eRe
Potential energy 643 Bio _
‘Latent heat 1060 Bu ents
Heat input
Heat to cooling water = 0.85 x (1667 - 600) MW = 907 MWe
ins _ke m 3)
eos
TIC
asu 70°F room
RS tORIS
254¢(140~70°F)
En
Evaporation rate = 907 x 10 KW x
1.23 Waterheaterjacket:
= 117 Bwhr
CEA
E
S99 KWhelyr
worth 599 kWhiyr x 0.08 $/kWhr = $47.92 yr
124. 60-wattincandescents vs 15-watt CFL:
Energy savings = (60 - 15 watts) x 9 kh
405 kW per CFL
Carbon savings = 405 kWh x 280 gC/KWh = 113,400 gC = 1134 kg C
1348502 = 1.134 kg SO2
Paniculate savings = 405 kWh x 0.14 ghkWh = 56:7 pariculate mater
‘$02 savings = 405 kWh x 2.8 g SO2/KWh:
Page 1.8
tal energy of Ki 643 Bro
Potential energy of 1 b at 5000 ft = 5000 fi eRe
Potential energy 643 Bio _
‘Latent heat 1060 Bu ents
Heat input
Heat to cooling water = 0.85 x (1667 - 600) MW = 907 MWe
ins _ke m 3)
eos
TIC
asu 70°F room
RS tORIS
254¢(140~70°F)
En
Evaporation rate = 907 x 10 KW x
1.23 Waterheaterjacket:
= 117 Bwhr
CEA
E
S99 KWhelyr
worth 599 kWhiyr x 0.08 $/kWhr = $47.92 yr
124. 60-wattincandescents vs 15-watt CFL:
Energy savings = (60 - 15 watts) x 9 kh
405 kW per CFL
Carbon savings = 405 kWh x 280 gC/KWh = 113,400 gC = 1134 kg C
1348502 = 1.134 kg SO2
Paniculate savings = 405 kWh x 0.14 ghkWh = 56:7 pariculate mater
‘$02 savings = 405 kWh x 2.8 g SO2/KWh:
Page 1.8
1.25 75-wattincandescents vs 18-watt CFL:
2. Electricity savings = (75 - 18 watts) x 10 khr = 570 kWh per CFL
STOKWh 3412 Btu
. Heatiotoplantsaved x10" Bu
b Heatinto plant an HD santo
06 Ibs $0, ton
SO, saved = DMI à 5.4510 Brox = 0.00162 tons
10° Buu into plant ig Be 2000 Ibs
©. Allowances = $800 / ton x 0.00162 ton / CFL = 51.30 per CFL
1.26 AL40% efficency, the power plant needs 2.5 kWh of heat input foreach | kWhrof
elccuicity delivered.
25KWihen
ThWhetectricity “KW
16005 20kg C
PELA
Carbon : EE 015 KC
2S5kWhheat 10% Klls
SO: TaWhelecticity "KW
0.77 g80,/ kWh
25kWh heat _, 10 Ws 36004 130350, 176, awn
TkWhelectrciy EW he 10"
NO,:
1.27 Mars with a peak wavelength of 13.2 ju
3x 10° mis
Br 7227 x 10° Be
E=hv=66 x 10% 1-5 x 227 x 10%/s = 15:10") /photon
Page 19
2. Electricity savings = (75 - 18 watts) x 10 khr = 570 kWh per CFL
STOKWh 3412 Btu
. Heatiotoplantsaved x10" Bu
b Heatinto plant an HD santo
06 Ibs $0, ton
SO, saved = DMI à 5.4510 Brox = 0.00162 tons
10° Buu into plant ig Be 2000 Ibs
©. Allowances = $800 / ton x 0.00162 ton / CFL = 51.30 per CFL
1.26 AL40% efficency, the power plant needs 2.5 kWh of heat input foreach | kWhrof
elccuicity delivered.
25KWihen
ThWhetectricity “KW
16005 20kg C
PELA
Carbon : EE 015 KC
2S5kWhheat 10% Klls
SO: TaWhelecticity "KW
0.77 g80,/ kWh
25kWh heat _, 10 Ws 36004 130350, 176, awn
TkWhelectrciy EW he 10"
NO,:
1.27 Mars with a peak wavelength of 13.2 ju
3x 10° mis
Br 7227 x 10° Be
E=hv=66 x 10% 1-5 x 227 x 10%/s = 15:10") /photon
Page 19
1.28
1.29
1.30
Solar constant Earth
Solar constant Mars
sE
Rearth + [ 150x10% km 7
st = Stan (RE) crois HE nv
15) over
event (2) 268 ve
src (2) a ve
ef]
STO ln 138 (22-+273) (5227) ]= 136 W
= oft) -(t)']
25.67x10°W Im? x2 m'{(80 +273) ~ 20+273)'] =925 W
Page 1.10
1.29
1.30
Solar constant Earth
Solar constant Mars
sE
Rearth + [ 150x10% km 7
st = Stan (RE) crois HE nv
15) over
event (2) 268 ve
src (2) a ve
ef]
STO ln 138 (22-+273) (5227) ]= 136 W
= oft) -(t)']
25.67x10°W Im? x2 m'{(80 +273) ~ 20+273)'] =925 W
Page 1.10
SOLUTIONS FOR CHAPTER 2
2.1 Combustion of propane:
a. C3Hg+ 502 — 3CO2 + 4H20
8 Smoles of Oz needed per mole propane
Imol CH, , SmolO; , 3280,
24H, * molCH, * molO.
e 100 8GHyx 163.6 20,
mol o A
0.x 2 x22.414x10° LE =0.255 m0,
4. 363.6 20, 22.14 o,
Re
213 m’ aie
3motCO. , 100SCHH, 22414107
© ol GH, "a4 mo CH,” mal
0.153 m° CO,
22 4C7HSN305 +2102 > 28C02 +10 H20+6N2
mol wt of TNT = 7x12 + Sxl + 3x14 + 6x16 = 227 g/mol
¿TNT , 21molO, , 3280,
LOST, 740 20,
ZT ginal * Amol TNT * molO, "80:
= M 3330
DOC ayy pete STEH 6 mo a
a A om D 16 mol
CHO.
so could be CH2O (mol wt=30, 100 low)
could be C2H4O2 (mol wt=60 g/mol) YES
24 mol wtof glucose (C6H1206) = 6x12 + 12x1 + 6x16= 180 g/mol
Deg essa, = aossom
25 mol wtof ethyl alcohol (CH3CH2OH) = 2x12 + 6x1 + Ix16=46 g/mol
043 Lalcohol | _790g_ mol
LL whiskey x DTA, TOs, Be = 738 moll.
IX whiskey“ Lalcohol * 468
3% mol
Pet
2.1 Combustion of propane:
a. C3Hg+ 502 — 3CO2 + 4H20
8 Smoles of Oz needed per mole propane
Imol CH, , SmolO; , 3280,
24H, * molCH, * molO.
e 100 8GHyx 163.6 20,
mol o A
0.x 2 x22.414x10° LE =0.255 m0,
4. 363.6 20, 22.14 o,
Re
213 m’ aie
3motCO. , 100SCHH, 22414107
© ol GH, "a4 mo CH,” mal
0.153 m° CO,
22 4C7HSN305 +2102 > 28C02 +10 H20+6N2
mol wt of TNT = 7x12 + Sxl + 3x14 + 6x16 = 227 g/mol
¿TNT , 21molO, , 3280,
LOST, 740 20,
ZT ginal * Amol TNT * molO, "80:
= M 3330
DOC ayy pete STEH 6 mo a
a A om D 16 mol
CHO.
so could be CH2O (mol wt=30, 100 low)
could be C2H4O2 (mol wt=60 g/mol) YES
24 mol wtof glucose (C6H1206) = 6x12 + 12x1 + 6x16= 180 g/mol
Deg essa, = aossom
25 mol wtof ethyl alcohol (CH3CH2OH) = 2x12 + 6x1 + Ix16=46 g/mol
043 Lalcohol | _790g_ mol
LL whiskey x DTA, TOs, Be = 738 moll.
IX whiskey“ Lalcohol * 468
3% mol
Pet
2.6 a. percentage chlorine:
en Go wate
a SSAA = 0.505 586%
Cane [2 + 213545 + 2x19
3545
22 (CHR): a 8 = ot
HCFO=22 (CAROD: Cl a
b. HCFC-22 substituting for CFC-11:
774g- 4108
Trés
47 = 47% reduction
2.7 Combustion of petroleum:
4C2H3 + 1102 >8C02 + 6H20
3x 107 kiye
9810 kg oil
TT id
a. world energy, cil equivalent
” 10° mol
art pmol = 698x107 on E x ML 22 58010 mo où
at 27 gimol 98 x 10 kg. WX We -2.58x10'* mol oil
E motel ZmelCO, , 44200. „Ike
emissions = 2.58% 10° mol ilx ZUSICO: x le
2 2 1G" mat sth mol oii mol CO, o n
=227x10" kg CO,
b. From example 23, ¡fall U.S. energy came from methane
4.1x10" kg CO,yr from methane a
4110" ke COL frommethant _ 310" lyr
82 x 10° ki/yrof energy from methane alone?
5x10? kg CO,
comparing that with the CO? released fall peroleum, yields
2.27x10° kg CO,/yrif bum oil
15x10” kg CO, /yrif bum CH,
St. that is, 51% more CO, if bum oil
Pg 22
en Go wate
a SSAA = 0.505 586%
Cane [2 + 213545 + 2x19
3545
22 (CHR): a 8 = ot
HCFO=22 (CAROD: Cl a
b. HCFC-22 substituting for CFC-11:
774g- 4108
Trés
47 = 47% reduction
2.7 Combustion of petroleum:
4C2H3 + 1102 >8C02 + 6H20
3x 107 kiye
9810 kg oil
TT id
a. world energy, cil equivalent
” 10° mol
art pmol = 698x107 on E x ML 22 58010 mo où
at 27 gimol 98 x 10 kg. WX We -2.58x10'* mol oil
E motel ZmelCO, , 44200. „Ike
emissions = 2.58% 10° mol ilx ZUSICO: x le
2 2 1G" mat sth mol oii mol CO, o n
=227x10" kg CO,
b. From example 23, ¡fall U.S. energy came from methane
4.1x10" kg CO,yr from methane a
4110" ke COL frommethant _ 310" lyr
82 x 10° ki/yrof energy from methane alone?
5x10? kg CO,
comparing that with the CO? released fall peroleum, yields
2.27x10° kg CO,/yrif bum oil
15x10” kg CO, /yrif bum CH,
St. that is, 51% more CO, if bum oil
Pg 22
2.8 a. propane: C3Hg +502 > 3CO2_ + 4H20
29
AS (0) 1385 42418) (2) net
42858) (I) gross
net AHO =
(241.8) + 3393.5) (103.8) = 2048.9 KIlmol
gross: AHO = 4x(-285.8) + 3(-393.5)- (108.8) =-2219.9 Kimol
b. n-butane: Ca +192 02 > 4COz + 5H20
Clas: © 43935) 5-241.) (net
52858) () gross
nets AHO = Sx(-241.8) + 4.3935) (124.7) =-26583 kfmol
gross: AHO = 5x(-2858) + (393.5) - (124.7) = -28783 kilmol
<: isobutane: CaHig +192 02 > 4002 + SHO
C318) © 43935) 5.2418) (9 net
52858) () gross
net: AHO = 5x(-241.8) + 4399.) - (131.6) = -2651.4 Kimo}
gross: AHO = Sx(-2858) + 4(-395.5) - (-131.6) = 2871.4 KJ/mol
4. ethanol (D: CzHSOH +302 > 2C02 + 3120
Ens © 39.5) 3(-241.8)(g) net
30.2858) (D gross
net: AH = 3x(-241.8) + 2(-393.5) - (277.6) = -12348 Kimo!
gross: AHO = 3x(-2858) + 2(-393.5) - (-277.6) = -1366.8 K/mol
e. methanol (D: CH3OH +32 02 > CO2 + 2H20
2389) © 3935) 2.241.) (g) net
22858) () gross
ner AH = 2x(-241.8) + (-393.5)- (238.6)
gross: AH = 2x(-2858) + (-393.5) - (-238.)
ethanol (I): C2HSOH +302 > 2C02 + 3120
(AH) © A385) 3.2858) (D gross
32858) + 2(-393.8) - AHO =-1370KI/mol
AHO = 2744 klimol
Pa23
29
AS (0) 1385 42418) (2) net
42858) (I) gross
net AHO =
(241.8) + 3393.5) (103.8) = 2048.9 KIlmol
gross: AHO = 4x(-285.8) + 3(-393.5)- (108.8) =-2219.9 Kimol
b. n-butane: Ca +192 02 > 4COz + 5H20
Clas: © 43935) 5-241.) (net
52858) () gross
nets AHO = Sx(-241.8) + 4.3935) (124.7) =-26583 kfmol
gross: AHO = 5x(-2858) + (393.5) - (124.7) = -28783 kilmol
<: isobutane: CaHig +192 02 > 4002 + SHO
C318) © 43935) 5.2418) (9 net
52858) () gross
net: AHO = 5x(-241.8) + 4399.) - (131.6) = -2651.4 Kimo}
gross: AHO = Sx(-2858) + 4(-395.5) - (-131.6) = 2871.4 KJ/mol
4. ethanol (D: CzHSOH +302 > 2C02 + 3120
Ens © 39.5) 3(-241.8)(g) net
30.2858) (D gross
net: AH = 3x(-241.8) + 2(-393.5) - (277.6) = -12348 Kimo!
gross: AHO = 3x(-2858) + 2(-393.5) - (-277.6) = -1366.8 K/mol
e. methanol (D: CH3OH +32 02 > CO2 + 2H20
2389) © 3935) 2.241.) (g) net
22858) () gross
ner AH = 2x(-241.8) + (-393.5)- (238.6)
gross: AH = 2x(-2858) + (-393.5) - (-238.)
ethanol (I): C2HSOH +302 > 2C02 + 3120
(AH) © A385) 3.2858) (D gross
32858) + 2(-393.8) - AHO =-1370KI/mol
AHO = 2744 klimol
Pa23
2.10 propane: C3Hg_ +502 > 3CO2_ + 4420
des 0) 339.5) 42858) (0) gross
2219.9 kJimol
gross: AHO = 4x(-2858) + 30393.) - 1035
2199 KL x MEL x 10008 = 50.452
Cr
241 a methanol): CHSOH +32 02 > CO2 + 2120
89 ) (393.5) 26-2858) 0) gross
‘AHO = 2x(-2858) + (393,5) (:238.6) = -726.5 kJ/mol
gross
mmol 10008 LG, BO
eyes EL x BA PTE
a 78 32g 221b gal
= 65,537Biul
o TB
b. ethanol (D: C2HS0H +302 > 2C02 + 3120.
an) © A393) 32858) () gross
gross: AHO = 3x(-2858) + 2393.5) - (-277.6) = -13668 Klfmal
mol, 10008, gl, Bu
= * agg 220 gal 105
Huy = 136682 84,492 Bru/gal
e. propane: C3Hg_ +502 > 3C02_ + 4H20.
dos (0) 3305) 4.2858) ( gross
gross: AHO = 4x(-285.8) + 3(-393.5)- (-103.8) = -2219.9 Kl/mol
mel 1000g pb Bi
ve 221998 x M OE AE Tass 9 Bu
HHY = 221 TIN
2.12 a methanol ): CHSOH +32 02 > COz + 210
(389) (3935) 22418) () net
net: AHO = 2x(-241.8) + (393.5) - (2386) = -638.5 KJlmol
N 10008 grid, Bu
LHV = 68. „10002 - sl
He Sn *328° 226 gal 1055K a
ethanol (0: CHSOH +302 > 2C02 + 3120
EPS © I 3624180 net
net: AH® = 3x(-241.8) + 2(-393.5) - (-277.6) = 12348 kJ/mol
Lay mas mol, 10008 660, Bu
xs x66 x 16,392 Big
sl 468 221b gal 1.055 kt ee
Pg24
des 0) 339.5) 42858) (0) gross
2219.9 kJimol
gross: AHO = 4x(-2858) + 30393.) - 1035
2199 KL x MEL x 10008 = 50.452
Cr
241 a methanol): CHSOH +32 02 > CO2 + 2120
89 ) (393.5) 26-2858) 0) gross
‘AHO = 2x(-2858) + (393,5) (:238.6) = -726.5 kJ/mol
gross
mmol 10008 LG, BO
eyes EL x BA PTE
a 78 32g 221b gal
= 65,537Biul
o TB
b. ethanol (D: C2HS0H +302 > 2C02 + 3120.
an) © A393) 32858) () gross
gross: AHO = 3x(-2858) + 2393.5) - (-277.6) = -13668 Klfmal
mol, 10008, gl, Bu
= * agg 220 gal 105
Huy = 136682 84,492 Bru/gal
e. propane: C3Hg_ +502 > 3C02_ + 4H20.
dos (0) 3305) 4.2858) ( gross
gross: AHO = 4x(-285.8) + 3(-393.5)- (-103.8) = -2219.9 Kl/mol
mel 1000g pb Bi
ve 221998 x M OE AE Tass 9 Bu
HHY = 221 TIN
2.12 a methanol ): CHSOH +32 02 > COz + 210
(389) (3935) 22418) () net
net: AHO = 2x(-241.8) + (393.5) - (2386) = -638.5 KJlmol
N 10008 grid, Bu
LHV = 68. „10002 - sl
He Sn *328° 226 gal 1055K a
ethanol (0: CHSOH +302 > 2C02 + 3120
EPS © I 3624180 net
net: AH® = 3x(-241.8) + 2(-393.5) - (-277.6) = 12348 kJ/mol
Lay mas mol, 10008 660, Bu
xs x66 x 16,392 Big
sl 468 221b gal 1.055 kt ee
Pg24
e. propane: C3H§ +502 > 3CO2_ + 4120
(1038) (0) 3385) 42418) (0 net
net: AHO = 4x(-241.8) + 3(-393.5) - (-103.8) = -2043.9 Kilmol
10005 4 tb, Bio
y
LAV =D po 2216 gi" LOSS
mal 448
82,057 Buugal
213 O3 + hv > O2 + OF
(1429) © (a
AHO = 438-142.9=295.1 kJ/mol (endothermic)
LIO x 10 KI mimo!
aH
As am
1.19 x 10* Kl. mimol
=0.403x10m =0.408um
295.1 mol
214 NO > NO +0
39) (90.4) (247.5) (enthalpies from Table 22)
AHO = 90.4+247.5-33.9= 304 klimol (endothermic)
1.19 x 10K - m/mol
ds Kar)
1.19 x 10° KI m/mol
=039x10%m = 0.35)
304 mol si
28 Ha > mec
2 x IR, md. ass x 10tmeuL = [Hr]
1 Tooome * 3645 HO
pH = -log [H*] =- log [68587 x 104] = 3.16
P25
(1038) (0) 3385) 42418) (0 net
net: AHO = 4x(-241.8) + 3(-393.5) - (-103.8) = -2043.9 Kilmol
10005 4 tb, Bio
y
LAV =D po 2216 gi" LOSS
mal 448
82,057 Buugal
213 O3 + hv > O2 + OF
(1429) © (a
AHO = 438-142.9=295.1 kJ/mol (endothermic)
LIO x 10 KI mimo!
aH
As am
1.19 x 10* Kl. mimol
=0.403x10m =0.408um
295.1 mol
214 NO > NO +0
39) (90.4) (247.5) (enthalpies from Table 22)
AHO = 90.4+247.5-33.9= 304 klimol (endothermic)
1.19 x 10K - m/mol
ds Kar)
1.19 x 10° KI m/mol
=039x10%m = 0.35)
304 mol si
28 Ha > mec
2 x IR, md. ass x 10tmeuL = [Hr]
1 Tooome * 3645 HO
pH = -log [H*] =- log [68587 x 104] = 3.16
P25
.16 {on ]=310° té
Em
L * Toon * 178
1.7810 moVL
eo E
(ue = 56x10" mat
pH = -log[H*]=-108(566x 10°7 1=625
.17 pH = -log(H*]=8.5 2 [H*] = 3.16 x 10% mov
S 10%
ER CE
CT ETS
.18 a. CH3COOH+ 202 > 2C02 + 2H20
‘Acetic acid mol wt = 2412 +4x1 +2x16 = 60 g/mol
mon TEA, By BOLAA. 20102 3280, 1000 915 ma
TT TN]
b. CoHSOH +302 > 2C02 +3 H20 ethanol mol wt =46 g/mol
me, g ml 3010, 3250, 100mE 67 cm,
mu AmO, 280,
oP L 1000mg gan” moleth molO, * g
0:
e. CéH1206 + 602 > 6C02 + 6H2O suerose mol wt= 180 g/mol
3, molme „SmdiO, 3240, 1000mg
E mole any
180 g suc “nok suc TO 8 Saar
P26
Em
L * Toon * 178
1.7810 moVL
eo E
(ue = 56x10" mat
pH = -log[H*]=-108(566x 10°7 1=625
.17 pH = -log(H*]=8.5 2 [H*] = 3.16 x 10% mov
S 10%
ER CE
CT ETS
.18 a. CH3COOH+ 202 > 2C02 + 2H20
‘Acetic acid mol wt = 2412 +4x1 +2x16 = 60 g/mol
mon TEA, By BOLAA. 20102 3280, 1000 915 ma
TT TN]
b. CoHSOH +302 > 2C02 +3 H20 ethanol mol wt =46 g/mol
me, g ml 3010, 3250, 100mE 67 cm,
mu AmO, 280,
oP L 1000mg gan” moleth molO, * g
0:
e. CéH1206 + 602 > 6C02 + 6H2O suerose mol wt= 180 g/mol
3, molme „SmdiO, 3240, 1000mg
E mole any
180 g suc “nok suc TO 8 Saar
P26
HOCI @ Ht + OCI K=29x 108
Joa
=29x10*
Hocı)
ee tt
modi TONE CIO
on pe 107
2.20 HS > Ht + HS K =086x 107
HS] 1 1 1
ESS) LES] Lo" OBR
ag ET en
1
Pins ao
107
pH=8 aaa 0.104
10”
221 AIPOs = ABY + PO, 1022 [AB+) POP],
TA] = [POs]. [POs] = 10711 mol/L.
mol wi PO4°- = 31+ 4x16= 95 g/mol
PO43* concentration = 10-11 mol/L x 95 g/mol x 1000 mg/g = 9.5 x 10-7 mg/L.
Pg27
Joa
=29x10*
Hocı)
ee tt
modi TONE CIO
on pe 107
2.20 HS > Ht + HS K =086x 107
HS] 1 1 1
ESS) LES] Lo" OBR
ag ET en
1
Pins ao
107
pH=8 aaa 0.104
10”
221 AIPOs = ABY + PO, 1022 [AB+) POP],
TA] = [POs]. [POs] = 10711 mol/L.
mol wi PO4°- = 31+ 4x16= 95 g/mol
PO43* concentration = 10-11 mol/L x 95 g/mol x 1000 mg/g = 9.5 x 10-7 mg/L.
Pg27
Onis 21% of air, so at sea level (1 atm), P O2 = 0.21 x À atm=021 atm
ALISOC, KH =0.0015236 for O2 (Table 2.4)
[02] = KH Pg =0.0015236 mol/L-atm x 0221 atm = 0.00031996 mol/L,
dissolved oxygen =3.1996 x 104 mol/L x32 g/mol x 1000 mp/e= 102 mg/L
2 115% 10-4 H= 1 - 1.15 x 10-4 x 2000 = 0.77 atm
AZ km elevation,
102) = Kit Ps
dissolved oxygen = 2.464 x 10-4 mo. x32 g/mol x 1000 me
10002464 mol.
(0015236 mol/L-atm x 0.21 x 0.77 atm =
=79 mg/L
Pco2 =2atm and KH =0.033363 mol/L-atm at 25°C (Table 2.4)
[CO2} = KH Pg = 0.083363 mol/L-atm x 2 atm = 0.066676 mot
on in g/L. = 0.066676 mol/L x 44 g/mol =2.9 g/L)
CO2 + H20 => H + HCOS"
ICO] „x 400? mot.
CARS
and (H*] = [HCO3"] + [OH] (2.43)
and [H+] (OH) = Kw = 1014 (2.19)
1,10%: (C04) , 107%
co; IO,
eon ry ey TT
TH*P = Ky (CO2) + 10-1
{Hs} = 0.000173. molt.
ph = log (0000173) = 376
[x
44107 x 0.066676 + 10-14
Pg28
ALISOC, KH =0.0015236 for O2 (Table 2.4)
[02] = KH Pg =0.0015236 mol/L-atm x 0221 atm = 0.00031996 mol/L,
dissolved oxygen =3.1996 x 104 mol/L x32 g/mol x 1000 mp/e= 102 mg/L
2 115% 10-4 H= 1 - 1.15 x 10-4 x 2000 = 0.77 atm
AZ km elevation,
102) = Kit Ps
dissolved oxygen = 2.464 x 10-4 mo. x32 g/mol x 1000 me
10002464 mol.
(0015236 mol/L-atm x 0.21 x 0.77 atm =
=79 mg/L
Pco2 =2atm and KH =0.033363 mol/L-atm at 25°C (Table 2.4)
[CO2} = KH Pg = 0.083363 mol/L-atm x 2 atm = 0.066676 mot
on in g/L. = 0.066676 mol/L x 44 g/mol =2.9 g/L)
CO2 + H20 => H + HCOS"
ICO] „x 400? mot.
CARS
and (H*] = [HCO3"] + [OH] (2.43)
and [H+] (OH) = Kw = 1014 (2.19)
1,10%: (C04) , 107%
co; IO,
eon ry ey TT
TH*P = Ky (CO2) + 10-1
{Hs} = 0.000173. molt.
ph = log (0000173) = 376
[x
44107 x 0.066676 + 10-14
Pg28
© 275 ppm: (C02) = KH Py =0.033363 moV/L-atm x 275x10Satm
17 x 10-6molL
using (2.46): 1H*12= Ky [CO2 (ag)] + 10-14
{HA} = 447 x 10-7 x 9.17 x 1076 + 10-14 = 4.11 x 10-12
[Ht] = 20x106
pH = lo 1 =- 19g 205 10) = 569
@ 600 ppm: [CO2)= 0.033363 moV/L-aten x 600x106 atm = 2.0 x10°5 mol
using (246): (H*}2=
47 x 107x2.0x 10-5+ 10-14 =8.95 x 10-12
[H+] = 299 x 106
pH = -log [H*] =- log (2.99 x 10-6) = 5.52
2.26 Begin by writing the full set of equations that must be satisfied:
WJnco, A
Ss AO? mL 037)
RE Km 460% mol 238)
[Too ak, e 4510 matt, (230)
[1] +2[ca**]=[100,]+2[00,*]+[om"] =»[Hco;
+[0m] charge balance
[COr(aa)) = KH Pg = 0033363 mol/Laim x 360510 atm = 12 x 10 SmolL
Solve forthe concentration of Ca2* using (2.37), (238), and (239):
| [cs
from the charge balance:
(H*] = [HCO3] + [OH] -2[Ca?*}
| Pg29
17 x 10-6molL
using (2.46): 1H*12= Ky [CO2 (ag)] + 10-14
{HA} = 447 x 10-7 x 9.17 x 1076 + 10-14 = 4.11 x 10-12
[Ht] = 20x106
pH = lo 1 =- 19g 205 10) = 569
@ 600 ppm: [CO2)= 0.033363 moV/L-aten x 600x106 atm = 2.0 x10°5 mol
using (246): (H*}2=
47 x 107x2.0x 10-5+ 10-14 =8.95 x 10-12
[H+] = 299 x 106
pH = -log [H*] =- log (2.99 x 10-6) = 5.52
2.26 Begin by writing the full set of equations that must be satisfied:
WJnco, A
Ss AO? mL 037)
RE Km 460% mol 238)
[Too ak, e 4510 matt, (230)
[1] +2[ca**]=[100,]+2[00,*]+[om"] =»[Hco;
+[0m] charge balance
[COr(aa)) = KH Pg = 0033363 mol/Laim x 360510 atm = 12 x 10 SmolL
Solve forthe concentration of Ca2* using (2.37), (238), and (239):
| [cs
from the charge balance:
(H*] = [HCO3] + [OH] -2[Ca?*}
| Pg29
[c0,] 10 2K, (HT
Rico)
sico quo ¿Ele
so (HT
2x4.57x107|
[HP 447107120" +10" A
mazo”
Bar
(e ae]
or (HP + 2.75 x 10-14 (H+? = 147 x 1025
Imagine pH>7 for example: (H*] < 10-7 then 2.75 x 10-14 (H+)? < 2.75x10°28
which is negligible compared to 1.47 x 1025 so we will ignore the [H*] term...
then (HB = 147x 1025 so [H*]=527 x 109
pit =-1og(527x 109) =83
227 à Calla
HOH
A a
CECI CE
EL HH
b. CH3CHCICH3
H Ci H
Pg 2.10
Rico)
sico quo ¿Ele
so (HT
2x4.57x107|
[HP 447107120" +10" A
mazo”
Bar
(e ae]
or (HP + 2.75 x 10-14 (H+? = 147 x 1025
Imagine pH>7 for example: (H*] < 10-7 then 2.75 x 10-14 (H+)? < 2.75x10°28
which is negligible compared to 1.47 x 1025 so we will ignore the [H*] term...
then (HB = 147x 1025 so [H*]=527 x 109
pit =-1og(527x 109) =83
227 à Calla
HOH
A a
CECI CE
EL HH
b. CH3CHCICH3
H Ci H
Pg 2.10
2:28 a. CH3CH2CH2CH2CH3 b. CH3CH2CHCH2
OR IE
e a
Ir ...
229 a. CH3CHCHCH3 b. CHSCHCICHQCH3 c. CH3CH2OH
HOH HOH Ha oH OH HH
111 RE 1
H-C-C=C-C-H H-C-C-C-C-H H-C-C-0H
1 1 IN] Da
H H HH HOH HOH
2.30 a. dichloromethane bo tichloromethane ©. 1,1«dichloroethylene
Î a cı cli
1 1 ba
I H-c-ci H-c-Cl c=c
Ë 1 I tt
! H ct ci H
4.trichlorofluoromethane _ «.1,1,2,2-1etrachloroethane f. o-dichlorobenzene
cl ci ci cl
il 1
i
F+C-CI H-C-C-H él
i
ci ci ci
i
|
N g, terachloroethene (PCE) h. dichlorfiuoromethane
| ci cl cl
to i
! Cec F-C-H
É rl Í
| ci cl cı
Pg 21
OR IE
e a
Ir ...
229 a. CH3CHCHCH3 b. CHSCHCICHQCH3 c. CH3CH2OH
HOH HOH Ha oH OH HH
111 RE 1
H-C-C=C-C-H H-C-C-C-C-H H-C-C-0H
1 1 IN] Da
H H HH HOH HOH
2.30 a. dichloromethane bo tichloromethane ©. 1,1«dichloroethylene
Î a cı cli
1 1 ba
I H-c-ci H-c-Cl c=c
Ë 1 I tt
! H ct ci H
4.trichlorofluoromethane _ «.1,1,2,2-1etrachloroethane f. o-dichlorobenzene
cl ci ci cl
il 1
i
F+C-CI H-C-C-H él
i
ci ci ci
i
|
N g, terachloroethene (PCE) h. dichlorfiuoromethane
| ci cl cl
to i
! Cec F-C-H
É rl Í
| ci cl cı
Pg 21
231 GX da + “Sy à2262 b=86
IX py
232 648 > 32g > 163 > Be > 4g > 2% > Ig
elapsed days: 60 120 180 240 300 360days
IX py
232 648 > 32g > 163 > Be > 4g > 2% > Ig
elapsed days: 60 120 180 240 300 360days
SOLUTIONS FOR CHAPTER 3
3.1 1 billion in 1850 growing to billion in 1975:
a. by doubling time: | billion — 2 billion — billion
means 2 doublings in 1975 - 1850 = 125 years.
125 yes
Faoublings
7. 70
ltr
a y
2.5 yrsdoubling
b. by formula:
A
1 o($)eoon =a
3.2 Taition from $1500 10 $20,000 in 1995 - 1962 = 33 yrs
2
ba In 25 yrs: N= Ne" = 20,000 € 25142,317/ yr
10
ABO) oras
N)._1 19
solos lat ann mt)
increasing at that rate from 1850 10 2000, the population would be
Se) 1.08 billion
3.4
Land = (land/food) x (food/keal) x (keal person) x (population)
@1% @0S% BOIS 15%
F=-1.040,540.14 1.5=1.1%/yr growth in land required
a. doubling time fortand needed = T, = 22 9% u yrs
Son
land land) ane sf ed en
el. (Ez)... (RS) ati o etant
Pe.31
3.1 1 billion in 1850 growing to billion in 1975:
a. by doubling time: | billion — 2 billion — billion
means 2 doublings in 1975 - 1850 = 125 years.
125 yes
Faoublings
7. 70
ltr
a y
2.5 yrsdoubling
b. by formula:
A
1 o($)eoon =a
3.2 Taition from $1500 10 $20,000 in 1995 - 1962 = 33 yrs
2
ba In 25 yrs: N= Ne" = 20,000 € 25142,317/ yr
10
ABO) oras
N)._1 19
solos lat ann mt)
increasing at that rate from 1850 10 2000, the population would be
Se) 1.08 billion
3.4
Land = (land/food) x (food/keal) x (keal person) x (population)
@1% @0S% BOIS 15%
F=-1.040,540.14 1.5=1.1%/yr growth in land required
a. doubling time fortand needed = T, = 22 9% u yrs
Son
land land) ane sf ed en
el. (Ez)... (RS) ati o etant
Pe.31
Population) x (Energy/person) x (Carbonleneray)
aise ese 21%
3.5 Carbon emissions
a. to double the current rate ol carbon emissions, at r= 1.5 + 1.5+ 1 =4%/y%
fa).
{EN np 21733 m
Ñ (à) array
in 1733 ya the pr capita energy demand will ave increased by
(Ener perso
(Energy (Person) ange @ wef 1.297 (up by = 30%)
(Energy! person) os
€. by then, total energy demand would have increased by
Energy =(Energlperson) x (population)
i at)
@15% (chat is, total growth @ 39/0)
rer an y gt eos
1.68 (up by 68%)
Es °
3.6 Starting with 5109 tomes Cy O 4% yr and total atmospheric C= 700x109 tonnes
a. Using (3.16) the time to emit 700 x 109 tonnes would be |
04
Lf sho OE
et ET
+, To double atmospheric C given that half of the emissions remain in he
Tinosphere would require total emissions of 700 x 2= 1400 billion tonnes:
Q 1, (004x1400x10° |)
Le (ea
Energy(kJ/yr) _ Carbon(kgC)
3.7 Carbon(ig/yr)= Population x EE); HOME
Cubou(es/r) = Popa person Energy(kI)
@06m @05% 03% rroul= 08% = 0.008
a. First find the carbon emission rate in 1990:
Co = 250x106 people x 320x106 kifperson x 15x10 KC
2x 1012 kgC
if this grows at 0.8% for 2020 - 1990 = 30 yes, emissions will be
aise ese 21%
3.5 Carbon emissions
a. to double the current rate ol carbon emissions, at r= 1.5 + 1.5+ 1 =4%/y%
fa).
{EN np 21733 m
Ñ (à) array
in 1733 ya the pr capita energy demand will ave increased by
(Ener perso
(Energy (Person) ange @ wef 1.297 (up by = 30%)
(Energy! person) os
€. by then, total energy demand would have increased by
Energy =(Energlperson) x (population)
i at)
@15% (chat is, total growth @ 39/0)
rer an y gt eos
1.68 (up by 68%)
Es °
3.6 Starting with 5109 tomes Cy O 4% yr and total atmospheric C= 700x109 tonnes
a. Using (3.16) the time to emit 700 x 109 tonnes would be |
04
Lf sho OE
et ET
+, To double atmospheric C given that half of the emissions remain in he
Tinosphere would require total emissions of 700 x 2= 1400 billion tonnes:
Q 1, (004x1400x10° |)
Le (ea
Energy(kJ/yr) _ Carbon(kgC)
3.7 Carbon(ig/yr)= Population x EE); HOME
Cubou(es/r) = Popa person Energy(kI)
@06m @05% 03% rroul= 08% = 0.008
a. First find the carbon emission rate in 1990:
Co = 250x106 people x 320x106 kifperson x 15x10 KC
2x 1012 kgC
if this grows at 0.8% for 2020 - 1990 = 30 yes, emissions will be
Ca Ce" = 12x10! keC lyre "1.525 x 10% kgCiyr
b. During those 30 years, the total emitted wll be (3.15):
Pafen_ 12 Gr
(er) 2 SO
es Door
08s) = 407 GC
e. Total energy demand in 2020 will be:
Enery=(Enggy/peon x (paper)
ET EN oral growth rate = 1.19 = O01
Energy (1980) = 320 x 106 (person) x 250 x 108 peopl
x 1016 yr
Energy (2020) = 8 x 1016 k/yre0.011 x 30 = 11.1 x 1016 k/yr
d. Percapita emissions of carbon in 2020 will be:
Ciperson(1990) = (Energy kilyr/person) x (Carbon/energy KgC/KI)
= 320x106 Kliyriperson x 15x10-SkgCiks = 4800 kaCiyr/person
growing at r=0.5% - 0.3% = 0.25blyr = 0.002)yr
C(2020) = C(1990) eft = 4800 kaC/yriperson e0.002/yr x 30yr = 5.1x10°kgCiyr
3.8 Current usage 2 million tons Cr/yr; reserves 800 million tons of chromium: r=2.6%lyr
292 41] tn 0026800000 |
uf) tm)
If resources are S reserves, the time to use them up would be
qe yl 2028252800810") 507 595
= ET
Note multiplying reserves x Sony increases te iene by a factor of 16
3.9 Gausian peaking 6 current te of milion tons eouro of billion sons
4000x110" tons
» ee ar 1m Monza)
‘To reach the maximum production rate, use (3.20):
cane eE rare
Pg. 33
b. During those 30 years, the total emitted wll be (3.15):
Pafen_ 12 Gr
(er) 2 SO
es Door
08s) = 407 GC
e. Total energy demand in 2020 will be:
Enery=(Enggy/peon x (paper)
ET EN oral growth rate = 1.19 = O01
Energy (1980) = 320 x 106 (person) x 250 x 108 peopl
x 1016 yr
Energy (2020) = 8 x 1016 k/yre0.011 x 30 = 11.1 x 1016 k/yr
d. Percapita emissions of carbon in 2020 will be:
Ciperson(1990) = (Energy kilyr/person) x (Carbon/energy KgC/KI)
= 320x106 Kliyriperson x 15x10-SkgCiks = 4800 kaCiyr/person
growing at r=0.5% - 0.3% = 0.25blyr = 0.002)yr
C(2020) = C(1990) eft = 4800 kaC/yriperson e0.002/yr x 30yr = 5.1x10°kgCiyr
3.8 Current usage 2 million tons Cr/yr; reserves 800 million tons of chromium: r=2.6%lyr
292 41] tn 0026800000 |
uf) tm)
If resources are S reserves, the time to use them up would be
qe yl 2028252800810") 507 595
= ET
Note multiplying reserves x Sony increases te iene by a factor of 16
3.9 Gausian peaking 6 current te of milion tons eouro of billion sons
4000x110" tons
» ee ar 1m Monza)
‘To reach the maximum production rate, use (3.20):
cane eE rare
Pg. 33
To consume about 80% of the resource corresponds to + 1.30:
180% = 2x 13. 6=2x 13 x 133 =346 yrs
3.10 Atcurrentrates Po it would take 100 yes to add Q tons of CFC tothe already existing
Qtons. Thatis,
Q
100 Po=Q or 10
“Then using (3.16), the time required to add those Q tons and double CFCs
1/2 1 it 849 yes
Tele GOO + 1) =549 y
3.11. Bismuth halflife is 4.85 days so using (3.8) the corresponding reaction rate K is
in2 12 __m2
Tri so Kai ng län
After 7 days the initial M
Na Ne 10307" "23.688
3.12 Reaction rate K = 0.2/éay, so from (3.8) the half-life is
3.13 Using the logistic curve (5.26) starting with No=63 billion in 2000, growing at
RowOOLSiye to a maximum of K=103 billion, first find growth rater
R, 0015
re = = 0.0386
NS)
K 103)
‘To reach billion, we need first to find t* the ime when size is K/2=5.15 billion:
Kon. 103
cas) ala)
We can use (3.27) to find when will it reach 9 billion:
11.7 yrs before 2000 (that is, 1988)
Pg. 3.4
180% = 2x 13. 6=2x 13 x 133 =346 yrs
3.10 Atcurrentrates Po it would take 100 yes to add Q tons of CFC tothe already existing
Qtons. Thatis,
Q
100 Po=Q or 10
“Then using (3.16), the time required to add those Q tons and double CFCs
1/2 1 it 849 yes
Tele GOO + 1) =549 y
3.11. Bismuth halflife is 4.85 days so using (3.8) the corresponding reaction rate K is
in2 12 __m2
Tri so Kai ng län
After 7 days the initial M
Na Ne 10307" "23.688
3.12 Reaction rate K = 0.2/éay, so from (3.8) the half-life is
3.13 Using the logistic curve (5.26) starting with No=63 billion in 2000, growing at
RowOOLSiye to a maximum of K=103 billion, first find growth rater
R, 0015
re = = 0.0386
NS)
K 103)
‘To reach billion, we need first to find t* the ime when size is K/2=5.15 billion:
Kon. 103
cas) ala)
We can use (3.27) to find when will it reach 9 billion:
11.7 yrs before 2000 (that is, 1988)
Pg. 3.4
u eR ——
1 (103,
Frl 384 yrs (that is 2038)
which is quite similar to Figure 3.20.
Similar to Problem 3.13 but starting with 3.65 billion in 1970 and 2.0% growth:
LR
Fer
1 Na
K
de
Projected out to 1995 (25 yes later) using Eq. 3.22:
9.4 yrs (19.4 + 1970 = 1989)
ES 103
ee
Ira
N 56 billion (actual was 5.7)
‘When No=100 the doubing ime en ls us fn the growth ate Ros
Rom 2 ye
Toi
With no growth constraints use (3.26).
R, 0.693,
rer =0711 yr
77,160
K 2000,
3. Max sustainable yield when population is half the camying capacity
N= 2000/2 = 1000 fish
using (3.29) the maximum yield is:
#K _ 0711 x 2000
4 4
b. Ifthe pond is kept at 1500 fish (instead of the optimum 1000), yield (3.21) is
=355 fishiyr
max yield
vied «en 711 x 1500 (115% 99) = 267 she
Pg. 35
1 (103,
Frl 384 yrs (that is 2038)
which is quite similar to Figure 3.20.
Similar to Problem 3.13 but starting with 3.65 billion in 1970 and 2.0% growth:
LR
Fer
1 Na
K
de
Projected out to 1995 (25 yes later) using Eq. 3.22:
9.4 yrs (19.4 + 1970 = 1989)
ES 103
ee
Ira
N 56 billion (actual was 5.7)
‘When No=100 the doubing ime en ls us fn the growth ate Ros
Rom 2 ye
Toi
With no growth constraints use (3.26).
R, 0.693,
rer =0711 yr
77,160
K 2000,
3. Max sustainable yield when population is half the camying capacity
N= 2000/2 = 1000 fish
using (3.29) the maximum yield is:
#K _ 0711 x 2000
4 4
b. Ifthe pond is kept at 1500 fish (instead of the optimum 1000), yield (3.21) is
=355 fishiyr
max yield
vied «en 711 x 1500 (115% 99) = 267 she
Pg. 35
3.16 Begin by finding the early growth rate r from (3.26)
sis 713000 (1 -30%o99) = 53 fistiyr
“This is less than the maximum sustainable yield of 710 fish found in Example 3.9.
3.17 At present, yield = dN/dt = 2000/yr; K= 10,000 fish; N= 4000; and we want
maximize yield. From (321)
ayy
a oe
CS) oof
UK 10,000,
“To maximize sustainable yield, the population should be allowed to grow to K/2=
5006 fish, at which point the yield would be:
IK _ 0.8333x10,000
4
= 2083 fishyr
3.18 India; 13/1000; infant mort=118 per 1000 live births:
a. bints=762x10% SE = 259 milion yr
1000
118 deaths
1000 births
infant deaths = 25.9x10° births x 06rillion/yr
total death
B
210 x = 091 mitiondeathsyr
62x10" x Ls «991 milion denke
tocioot eas hata ne» 32% 03006318
En
10 . 10dsaths á
infant deaths @12 = 259x10* bins x HOSS = 0.26milioniyr
® na 1000 ds 1000 births eo
“avoidable deaths” = 3.06 0.26 M = 28 million/yr
e. annual increase = N(b-<) 16 ilion/ye
Pe.36
sis 713000 (1 -30%o99) = 53 fistiyr
“This is less than the maximum sustainable yield of 710 fish found in Example 3.9.
3.17 At present, yield = dN/dt = 2000/yr; K= 10,000 fish; N= 4000; and we want
maximize yield. From (321)
ayy
a oe
CS) oof
UK 10,000,
“To maximize sustainable yield, the population should be allowed to grow to K/2=
5006 fish, at which point the yield would be:
IK _ 0.8333x10,000
4
= 2083 fishyr
3.18 India; 13/1000; infant mort=118 per 1000 live births:
a. bints=762x10% SE = 259 milion yr
1000
118 deaths
1000 births
infant deaths = 25.9x10° births x 06rillion/yr
total death
B
210 x = 091 mitiondeathsyr
62x10" x Ls «991 milion denke
tocioot eas hata ne» 32% 03006318
En
10 . 10dsaths á
infant deaths @12 = 259x10* bins x HOSS = 0.26milioniyr
® na 1000 ds 1000 births eo
“avoidable deaths” = 3.06 0.26 M = 28 million/yr
e. annual increase = N(b-<) 16 ilion/ye
Pe.36
74 per 1000 live births:
10 million Iyr
74 deaths
1000 bins
27.0x10° births x 22.0 millionlyr
9
total deaths = 931810" x ¡2 = 838 million deathsyr
1000 is
20
fraction of deaths that are infants = HE
fraction of deaths hat =
10 10 deaths
infant deaths =27.0x10° births x TOSSES
» infant deaths O; 055 = 27.0x10" bins x og is
"avoidable deaths" =
239 = 24%
0.27 million yr
.0-027M = 173 million/ye
(29 9
©. annual increase = N(b-d) =931x10'| 22
N NO 5-7
Notice even though birth rates are down, population is growing faster han it did in
1985 (Problem 3.18).
3.20 Replacement level fertility starting att=0:
so-7 [Mm LL am au
25-49 se Er am au
0-24 [TM zu au EN
O SM 4-25 BW t-50 SN 75 m
3.21 Same as Prob. 3.20 at beginning, but TFR=4 for 25yrs, and 20% death at 50:
sn Là vere <a an
sl 2 children. am
2-2 a Lu E
1-0 6M 1-25 106M
50-74 IL 2.40 (aux 0.8) m 4.M (6M x 0.8)
25-49 om sm
0-24 Sm (M x 2) SM (M x 2)
CCE 1275 Em
3.22 Since r=3.5%, the doubling time is Ta = 70/r = 703.5 = 20 yrs, So in 20 years the
population will have doubled from 290,000 to 580,000. The bottom of he pyramid
must Bethe amount that will make the total equal to 580,000:
Pg.37
10 million Iyr
74 deaths
1000 bins
27.0x10° births x 22.0 millionlyr
9
total deaths = 931810" x ¡2 = 838 million deathsyr
1000 is
20
fraction of deaths that are infants = HE
fraction of deaths hat =
10 10 deaths
infant deaths =27.0x10° births x TOSSES
» infant deaths O; 055 = 27.0x10" bins x og is
"avoidable deaths" =
239 = 24%
0.27 million yr
.0-027M = 173 million/ye
(29 9
©. annual increase = N(b-d) =931x10'| 22
N NO 5-7
Notice even though birth rates are down, population is growing faster han it did in
1985 (Problem 3.18).
3.20 Replacement level fertility starting att=0:
so-7 [Mm LL am au
25-49 se Er am au
0-24 [TM zu au EN
O SM 4-25 BW t-50 SN 75 m
3.21 Same as Prob. 3.20 at beginning, but TFR=4 for 25yrs, and 20% death at 50:
sn Là vere <a an
sl 2 children. am
2-2 a Lu E
1-0 6M 1-25 106M
50-74 IL 2.40 (aux 0.8) m 4.M (6M x 0.8)
25-49 om sm
0-24 Sm (M x 2) SM (M x 2)
CCE 1275 Em
3.22 Since r=3.5%, the doubling time is Ta = 70/r = 703.5 = 20 yrs, So in 20 years the
population will have doubled from 290,000 to 580,000. The bottom of he pyramid
must Bethe amount that will make the total equal to 580,000:
Pg.37
40-59 10,000 80,000
20-59 80.000 200,000
0-18 200,000 7
o 230,000 20 360,000
So the bottom of the pyramid must have $80,000 - (80.000+200,000) = 300,000
people. For 200,000 people to have had 300,000 births means on average each
aon in the 0-19 category had 1.5 children, or each woman had 3.0 children, That
IS TFR = 3.0
3.23 Sample calculations for 1990:
ages 0-9 in 1990 = P(1990) = b10P10(1980) + b20P20(1980)
= 0.25 x 224M + 0.25 x 182 M=101.5M
Le =
Lo
ages 10-19: P (1990) = P1980) EE =235M x 0957 = 2249
similarly, 20(1990) 221M
P30(1990) (784M
P30(1990) 1195M
Pso(1990) 878M
P60(1950) STOM
P70(1990) 266M
P80(1990) 26M |
3.24 Usi
spreadsheet makes it straightforward
Problem 324 Chia Age nue with EE Fa 1
T T
“=| Fraction [irths_paiPopultionFopulation Population Pop stop on PopsTation
iiierval Surviving! Person | 1980 | 1990 | 2000 | 2010 | 2020 | 2030
0-5 10357 ol 2351 101,501 111.501 79.781 50.64] 45.42
o 1510387 Dsl 2241 224.801 37.141 108.701 76351 48.47|
20-28] 0,98 0.28] 182] 221.081 221.971 95.871 10531 7535) |
30-38 0.964] ‘OL 124] 178.36] 216.671 217.53] 93.96] 108.21
20-39] 0.924 0] 9519.54] 17164] 208.871 209.70] 90.57
50-501 0.826] al sl 87.781 m04s| 158.871 192.991 193.76
(5069 10.633] 0! 421 56.901 72,511 91.23] 131.23] 159.01
70-731 0316| 0! 24126801 36.081 45.001 5775) 83.07
[601 ol ol 6 7,56] 640] 11401 14,501 18.25]
Hot T0011.1024.321 1046.64] 1016151 932.431 817.51
Pg, 38
20-59 80.000 200,000
0-18 200,000 7
o 230,000 20 360,000
So the bottom of the pyramid must have $80,000 - (80.000+200,000) = 300,000
people. For 200,000 people to have had 300,000 births means on average each
aon in the 0-19 category had 1.5 children, or each woman had 3.0 children, That
IS TFR = 3.0
3.23 Sample calculations for 1990:
ages 0-9 in 1990 = P(1990) = b10P10(1980) + b20P20(1980)
= 0.25 x 224M + 0.25 x 182 M=101.5M
Le =
Lo
ages 10-19: P (1990) = P1980) EE =235M x 0957 = 2249
similarly, 20(1990) 221M
P30(1990) (784M
P30(1990) 1195M
Pso(1990) 878M
P60(1950) STOM
P70(1990) 266M
P80(1990) 26M |
3.24 Usi
spreadsheet makes it straightforward
Problem 324 Chia Age nue with EE Fa 1
T T
“=| Fraction [irths_paiPopultionFopulation Population Pop stop on PopsTation
iiierval Surviving! Person | 1980 | 1990 | 2000 | 2010 | 2020 | 2030
0-5 10357 ol 2351 101,501 111.501 79.781 50.64] 45.42
o 1510387 Dsl 2241 224.801 37.141 108.701 76351 48.47|
20-28] 0,98 0.28] 182] 221.081 221.971 95.871 10531 7535) |
30-38 0.964] ‘OL 124] 178.36] 216.671 217.53] 93.96] 108.21
20-39] 0.924 0] 9519.54] 17164] 208.871 209.70] 90.57
50-501 0.826] al sl 87.781 m04s| 158.871 192.991 193.76
(5069 10.633] 0! 421 56.901 72,511 91.23] 131.23] 159.01
70-731 0316| 0! 24126801 36.081 45.001 5775) 83.07
[601 ol ol 6 7,56] 640] 11401 14,501 18.25]
Hot T0011.1024.321 1046.64] 1016151 932.431 817.51
Pg, 38
3
25 Now delay binhs by one 10-ye interval:
China Age Structure with 1-chid family, delayed bith: 1
Proven 32:
1 = i
RE | Fraction Bis “per Poputation FopulationPopulstion PopulationPopulation Population]
Tnterval Survivingl Person | 1980 { 1990 | 2000 | 2010 | 2020 | 2030
0-9 | 0.57 GI 2557680) 99.86] 109.681 72.45| 4128]
Ho = 150.887; Of 2241 224.801 73.211 95.571 104.54[ 69.35)
20-25 {0.98 0.25 182) 221.091 221.971 72.26] 94.331 10358]
150-301 8.864] os] 1741 17636] 216.67 217.531 7081] 92.44]
[30 - 49 [0.924 ‘OL 98) 119.54] 171,54] 208.871 209.70] _ 88.26)
50-501 0.828 SL 8887 .78| 110.451 158.87] 192.99] 193.76
[6969 | 0.633 ol 427 56.99] 7251] 91.23] 131.23] 159.41
70: 7010316) 024] 26:50] 36.0Bl 45.001 57.75| 83.91]
[80-1 0 0 el 7,58 8.40] 11.40] 14.50] 18.25]
front Ooi} 888.32] 1011.00] To1ı.zel s48.71| 820.39]
“The peak in Prob.3.24 was 1047 million: with delayed births the peak drops to 1011 million.
3.26 Using the more realistic 2-child per family birth rates gives:
Problem 3267 China Age Sisto ON Tarai Y I T
a 7 1 —
(Population PopulationPopulation Population]
2000 | 2010 | 2020 | ~2030
mi E
RE Fraction Birchs pero;
nal [Surviving| Person
0-5 | 0.957 al 182.00) 221.09] 221.971 171.911 208.83]
10-187 0.097] a 224.90) 174.17| 211-88) 212.431 164.52]
20-291 0.98 a 221-09] 221.97] 171.01] 208.83| 209.67
30-301 6.864] a 178.361 218.67] 217.531 168.471 204.68]
[3049 | 0.924 of 119.54] 171.94] 208.87| 209.701 162.41
50 5910.28 9 87.78 110.45) 158.871 192,98] 193.781
60-691 0.633 0 5639] 7251 91.23] 131.231 158.41
70-791 0316) 9 26.50] 36.08] 45.90] _ $7.75| 83.07]
E 6 0 758] _ 8.40 11.40} 14,501 _ 18.25
fon TOOT! 1104. 8211733.271 1339.28 1367-811 1404.57)
A A
Pg.39
25 Now delay binhs by one 10-ye interval:
China Age Structure with 1-chid family, delayed bith: 1
Proven 32:
1 = i
RE | Fraction Bis “per Poputation FopulationPopulstion PopulationPopulation Population]
Tnterval Survivingl Person | 1980 { 1990 | 2000 | 2010 | 2020 | 2030
0-9 | 0.57 GI 2557680) 99.86] 109.681 72.45| 4128]
Ho = 150.887; Of 2241 224.801 73.211 95.571 104.54[ 69.35)
20-25 {0.98 0.25 182) 221.091 221.971 72.26] 94.331 10358]
150-301 8.864] os] 1741 17636] 216.67 217.531 7081] 92.44]
[30 - 49 [0.924 ‘OL 98) 119.54] 171,54] 208.871 209.70] _ 88.26)
50-501 0.828 SL 8887 .78| 110.451 158.87] 192.99] 193.76
[6969 | 0.633 ol 427 56.99] 7251] 91.23] 131.23] 159.41
70: 7010316) 024] 26:50] 36.0Bl 45.001 57.75| 83.91]
[80-1 0 0 el 7,58 8.40] 11.40] 14.50] 18.25]
front Ooi} 888.32] 1011.00] To1ı.zel s48.71| 820.39]
“The peak in Prob.3.24 was 1047 million: with delayed births the peak drops to 1011 million.
3.26 Using the more realistic 2-child per family birth rates gives:
Problem 3267 China Age Sisto ON Tarai Y I T
a 7 1 —
(Population PopulationPopulation Population]
2000 | 2010 | 2020 | ~2030
mi E
RE Fraction Birchs pero;
nal [Surviving| Person
0-5 | 0.957 al 182.00) 221.09] 221.971 171.911 208.83]
10-187 0.097] a 224.90) 174.17| 211-88) 212.431 164.52]
20-291 0.98 a 221-09] 221.97] 171.01] 208.83| 209.67
30-301 6.864] a 178.361 218.67] 217.531 168.471 204.68]
[3049 | 0.924 of 119.54] 171.94] 208.87| 209.701 162.41
50 5910.28 9 87.78 110.45) 158.871 192,98] 193.781
60-691 0.633 0 5639] 7251 91.23] 131.231 158.41
70-791 0316) 9 26.50] 36.08] 45.90] _ $7.75| 83.07]
E 6 0 758] _ 8.40 11.40} 14,501 _ 18.25
fon TOOT! 1104. 8211733.271 1339.28 1367-811 1404.57)
A A
Pg.39
SOLUTIONS FOR CHAPTER 4
4.1 From the slope ofthe figure, potency = 0.001/0.1(mg/kg-) = 0.01 (mgkg-0)1
coy = Lomein’ 1200/48
70 kg
1000285 mg/kg
Risk = CDI x Potency = 0.000285 mg/kg -d x 0.01 (mg/kg~d)"=2.9x10%
4.2 02 ppb of PCB:
u coy = LAND me L421 15350 IO! a 35.10 mg à
70 kg x 365 diye x 70 yr
b. Risk = CDI x Potency = 235x106 mg/kg-d x 7.7 (mg/kg-d)-! = 18.1 x 106
10°people x 18.1x10° eancerperson - fe _ 9.96 cancenyr
€. Extra cancerelyr =
= a TS
193 deaths/yr
100.000 people
adding 0.26 cancers per year would not be detectable.
4.3 Ratdata:
4. Expected cancerrate = x 10° people =1930 deaths/yr
lol
b. Attributable risk = aod
ad _ 303290
e Odés >
rer er
Allee indicator are consistent witha relationship between exposure and risk.
4.4 Fillingin the matrix
o
E
Pg. 41
4.1 From the slope ofthe figure, potency = 0.001/0.1(mg/kg-) = 0.01 (mgkg-0)1
coy = Lomein’ 1200/48
70 kg
1000285 mg/kg
Risk = CDI x Potency = 0.000285 mg/kg -d x 0.01 (mg/kg~d)"=2.9x10%
4.2 02 ppb of PCB:
u coy = LAND me L421 15350 IO! a 35.10 mg à
70 kg x 365 diye x 70 yr
b. Risk = CDI x Potency = 235x106 mg/kg-d x 7.7 (mg/kg-d)-! = 18.1 x 106
10°people x 18.1x10° eancerperson - fe _ 9.96 cancenyr
€. Extra cancerelyr =
= a TS
193 deaths/yr
100.000 people
adding 0.26 cancers per year would not be detectable.
4.3 Ratdata:
4. Expected cancerrate = x 10° people =1930 deaths/yr
lol
b. Attributable risk = aod
ad _ 303290
e Odés >
rer er
Allee indicator are consistent witha relationship between exposure and risk.
4.4 Fillingin the matrix
o
E
Pg. 41
4.5
4.6
4.7
a. Relative risk
b. Attibutable ris
ad
e. Oddsratio == = 2.58
Odds ratio = SS = SF =
70 kg individual, 2L/day, 0.1 mg/L of 1,1-dichloroethylene. 20 years:
average dose during exposure _ 2L/d x0.1mglL /70kg
RD 0.009 mg/kg-d
Id x 0.1 mg/L x 20 yes a
be emp 2L1dx0.1 mg/L x 20 yrs x 365 dy
TO ka x 70 yrs x 365 diye
gx 70y i
a. Hazard quotient = =032
0.0008ISmg/xg-é
Risk
0.58 (mg/l =47x 104
DI x Poteney = 0.000816 mg
e. Drinking the water or 30 years instead of 20 doesn't change the Hazard Quotient
Sine thats based ony om the period of ime when he individuals exposed, The
‘cancer risk does change, however:
2LI4x0.1 mall. x 30 yrs x365 diyr
col
TO kg x 70 yrs x 365 yr
= 0.00122 me/kg-d
Risk =
100122 mg/kg-d x 0.58 (mglkg-4)1 = 7.1 x 104
DWEL is the concentration of pollutant that leads toa 10:6 risk,
Risk __10% m
= AEG 139x107 mg/kg=d, and using the definition of
core es = 133010" mag ds and sing te deinen of CD:
ms/L)x2L 1 a
= Ame/L)x2LIE 333x10*mg/kg=a, solving for
cor BL 00010 mg/Kg —á, saving for
mg/L) = DWEL = 70 x 1339x104/2=4,67x 103 myL=5ugL
« _ 2L/4x001 mglL,
Risk = CDI x Potency = 10° 2 PTE potency
so, Poteney = TEA a 610 (mg/kg ay"
0, Potency = SABA 03. 5x10™(ma hs a)
Pe. 42
4.6
4.7
a. Relative risk
b. Attibutable ris
ad
e. Oddsratio == = 2.58
Odds ratio = SS = SF =
70 kg individual, 2L/day, 0.1 mg/L of 1,1-dichloroethylene. 20 years:
average dose during exposure _ 2L/d x0.1mglL /70kg
RD 0.009 mg/kg-d
Id x 0.1 mg/L x 20 yes a
be emp 2L1dx0.1 mg/L x 20 yrs x 365 dy
TO ka x 70 yrs x 365 diye
gx 70y i
a. Hazard quotient = =032
0.0008ISmg/xg-é
Risk
0.58 (mg/l =47x 104
DI x Poteney = 0.000816 mg
e. Drinking the water or 30 years instead of 20 doesn't change the Hazard Quotient
Sine thats based ony om the period of ime when he individuals exposed, The
‘cancer risk does change, however:
2LI4x0.1 mall. x 30 yrs x365 diyr
col
TO kg x 70 yrs x 365 yr
= 0.00122 me/kg-d
Risk =
100122 mg/kg-d x 0.58 (mglkg-4)1 = 7.1 x 104
DWEL is the concentration of pollutant that leads toa 10:6 risk,
Risk __10% m
= AEG 139x107 mg/kg=d, and using the definition of
core es = 133010" mag ds and sing te deinen of CD:
ms/L)x2L 1 a
= Ame/L)x2LIE 333x10*mg/kg=a, solving for
cor BL 00010 mg/Kg —á, saving for
mg/L) = DWEL = 70 x 1339x104/2=4,67x 103 myL=5ugL
« _ 2L/4x001 mglL,
Risk = CDI x Potency = 10° 2 PTE potency
so, Poteney = TEA a 610 (mg/kg ay"
0, Potency = SABA 03. 5x10™(ma hs a)
Pe. 42
4.8 dioxin standard =3x10°8 mg/L. using EPA suggested exposure factors (Table 410)
DL/d x 3x10%mg/ Lx350dlyr x 30 yr
en 70 kg x 365 dlyr x 70 yr
5107 "mg/kg -d
Risk = CDI x Potency = 3510-10 mg x 1.566105 mg
4.9 Tetrachloroethylene standard =0.005 mg/L.
Risk =CDI x Potency
2UA x 0.005 mg/L x350 dye X30 Ne 5 110% mg/kg ay! =3.0x10*
70 Kg x 365 diy x 70 ye
4.10 Radiation poteney = 1 cancer death per 8000 person-+ems:
Cancer
b.
Risk = Amremiflight x N lights x
1 cancer death
cancer death; 260,10" people x 0.130 rem/yr =4.2810deaths /yr
3000 person rms "26010 poc di des ly
Leancer
5000 vems x 10 mem rem
= cross country flights
(agrees with Table 43)
au
Leançerdeath
Denver risk 20.12 rem/yr x 70 ye x Acancercesth,
i STO 8000 rems
Sealevel isk = 004 remfyr x 70 yrx Loera
8000 rem
b. Denver deaths due to cosmic radiation exposure:
OZ remiperson- ye x 0.510 people x death 4.5 9 athuyr
8000 people -rems
193 desthsiyr
ed = 05x] je x EEE 965 annual cancer deaths
Expected = 0.5x10 people x ¡py ogg peas © 65 annual cancer death
1 death
€. Incremental risk =10~ = (0.12 rem/yr -0.04 rem/yr) x N ye.
ment « OO remyt) x N yex Sy
800010
008
20.1 yr (Table 43 suggests 2 months)
Pg. 43
DL/d x 3x10%mg/ Lx350dlyr x 30 yr
en 70 kg x 365 dlyr x 70 yr
5107 "mg/kg -d
Risk = CDI x Potency = 3510-10 mg x 1.566105 mg
4.9 Tetrachloroethylene standard =0.005 mg/L.
Risk =CDI x Potency
2UA x 0.005 mg/L x350 dye X30 Ne 5 110% mg/kg ay! =3.0x10*
70 Kg x 365 diy x 70 ye
4.10 Radiation poteney = 1 cancer death per 8000 person-+ems:
Cancer
b.
Risk = Amremiflight x N lights x
1 cancer death
cancer death; 260,10" people x 0.130 rem/yr =4.2810deaths /yr
3000 person rms "26010 poc di des ly
Leancer
5000 vems x 10 mem rem
= cross country flights
(agrees with Table 43)
au
Leançerdeath
Denver risk 20.12 rem/yr x 70 ye x Acancercesth,
i STO 8000 rems
Sealevel isk = 004 remfyr x 70 yrx Loera
8000 rem
b. Denver deaths due to cosmic radiation exposure:
OZ remiperson- ye x 0.510 people x death 4.5 9 athuyr
8000 people -rems
193 desthsiyr
ed = 05x] je x EEE 965 annual cancer deaths
Expected = 0.5x10 people x ¡py ogg peas © 65 annual cancer death
1 death
€. Incremental risk =10~ = (0.12 rem/yr -0.04 rem/yr) x N ye.
ment « OO remyt) x N yex Sy
800010
008
20.1 yr (Table 43 suggests 2 months)
Pg. 43
1 death
4. 260%10" people x 0.080 rem x o a
1300 deathsly +
4.12 Radon exposure of 1.5 piC/L equivalent to 400 mremlyr (04 rem/yt):
a. O4remiyrx me x 260x10" people = 13,000 cancerdeaths/yr
Are Leancer _,_TOYE „3.5410 cancer lifetime.
de Risk = À 2000 person cn” Time
4.13 For75 million people exposed to 0.4 rem of radiation:
Leancer x
10015 Sap person ars © 75610" people
a. extra cancer deaths
b. normal cancerdeaths = 022 x 75x10" people =16.5x10'
4.14 Living within 5 miles of a reactor for 50 yrs--what mrem/yr exposun
„_.Xrem Leancerdeath
19% = AE x oye ==
09x10"
= 0.16810"
m yr= 0.16 meemiye
50
4.15 Concentrations yielding acceptable risks
a. benzene, oral, 10°5 risk, potency = 2.9x10-2 (mg/kg-d)!
Ri
CDI x Poteney
2L/4 x € mg x350 diye x30 yr, 2.9K
70 Ka x 365 diye x 70 yr mgikg-4
10°
10°°x70x365x70
+= 0.08 mg/l.
2x 350 x 30 x 2.9x10° aa
b. trichloroethylene in air, risk 10-6, inhalation potency 13x102:
20m /4 x € mg/m? x350 d/yrx30 yr , 13x10°
70 kg x 365dIyrx TO” mekg-d
__10*x70x365x70
= 20x350% 30 x 13x10"
=6.6x10* my
e. benzene in air, risk 10°5, potency 2.9x10°2:
Pg. 44
4. 260%10" people x 0.080 rem x o a
1300 deathsly +
4.12 Radon exposure of 1.5 piC/L equivalent to 400 mremlyr (04 rem/yt):
a. O4remiyrx me x 260x10" people = 13,000 cancerdeaths/yr
Are Leancer _,_TOYE „3.5410 cancer lifetime.
de Risk = À 2000 person cn” Time
4.13 For75 million people exposed to 0.4 rem of radiation:
Leancer x
10015 Sap person ars © 75610" people
a. extra cancer deaths
b. normal cancerdeaths = 022 x 75x10" people =16.5x10'
4.14 Living within 5 miles of a reactor for 50 yrs--what mrem/yr exposun
„_.Xrem Leancerdeath
19% = AE x oye ==
09x10"
= 0.16810"
m yr= 0.16 meemiye
50
4.15 Concentrations yielding acceptable risks
a. benzene, oral, 10°5 risk, potency = 2.9x10-2 (mg/kg-d)!
Ri
CDI x Poteney
2L/4 x € mg x350 diye x30 yr, 2.9K
70 Ka x 365 diye x 70 yr mgikg-4
10°
10°°x70x365x70
+= 0.08 mg/l.
2x 350 x 30 x 2.9x10° aa
b. trichloroethylene in air, risk 10-6, inhalation potency 13x102:
20m /4 x € mg/m? x350 d/yrx30 yr , 13x10°
70 kg x 365dIyrx TO” mekg-d
__10*x70x365x70
= 20x350% 30 x 13x10"
=6.6x10* my
e. benzene in air, risk 10°5, potency 2.9x10°2:
Pg. 44
jo 22m dx € mym x350 elyex30 yr , 23x10°
10% x 70 x 365 x 70
TT TD
4, vinyl chloride in water, risk 10-4, potency 23:
= 2L/dxC mg/L x350d/yrx30yt , 23
"TO kg x 365 dlyr x 70 yr inglkg-d
19x70 x365 x 70
22350 x30%23
23710 mg/L
4.16 Trichloroethylene in an industrial facility; risk 10-4:
Risk = CDI x Potency
à _ 20m dx € mg/m? 1250 dyrx2Syr „ 13x10°
TO kg x 365 dyn TO ye meng-d
10 x 70 x365x70
20 x 260 x 25 x 13x10"
10
11 mm?
TO Ka x 365 diye x 70 yr ngkg-d
To convert to ppm, we need the molecular weight of CACI3H, which is
2x12 +3x35.5 +1 = 131.5 g/mol. From (1.8):
24465 x C(mg/m®) _ 24.465 x 0.11
mol wt 1215, ee
4.17 Benzene in fish = C (mg/L) in river x BCF (L/kg): From Tabl
Risk = CDI x Potency
yo = EMB x 52 Lkg x 0.054 kg/d x 350 diye x30 yr „
le 4.12, BCI
2.9x10°
70 kg x 365 dlyrx 70 ye mgikg-a
10 x 70 x 365 x 70
32x 0.054 x 350 x 30.x2.9x10°
0.021 mg/L.
4.18 DDT in fish=C (mg/L) x BCF (kg); from Table 4.12, BCF
Rist
CDI x Potency
0020 ml x 54000 Likg x0.002 kg/d, _ 0.34
Oke ~ * myks-d
4.19 Hazard index = Sum of the hazard quotients:
Risk
Pg 45
54000 Like,
10% x 70 x 365 x 70
TT TD
4, vinyl chloride in water, risk 10-4, potency 23:
= 2L/dxC mg/L x350d/yrx30yt , 23
"TO kg x 365 dlyr x 70 yr inglkg-d
19x70 x365 x 70
22350 x30%23
23710 mg/L
4.16 Trichloroethylene in an industrial facility; risk 10-4:
Risk = CDI x Potency
à _ 20m dx € mg/m? 1250 dyrx2Syr „ 13x10°
TO kg x 365 dyn TO ye meng-d
10 x 70 x365x70
20 x 260 x 25 x 13x10"
10
11 mm?
TO Ka x 365 diye x 70 yr ngkg-d
To convert to ppm, we need the molecular weight of CACI3H, which is
2x12 +3x35.5 +1 = 131.5 g/mol. From (1.8):
24465 x C(mg/m®) _ 24.465 x 0.11
mol wt 1215, ee
4.17 Benzene in fish = C (mg/L) in river x BCF (L/kg): From Tabl
Risk = CDI x Potency
yo = EMB x 52 Lkg x 0.054 kg/d x 350 diye x30 yr „
le 4.12, BCI
2.9x10°
70 kg x 365 dlyrx 70 ye mgikg-a
10 x 70 x 365 x 70
32x 0.054 x 350 x 30.x2.9x10°
0.021 mg/L.
4.18 DDT in fish=C (mg/L) x BCF (kg); from Table 4.12, BCF
Rist
CDI x Potency
0020 ml x 54000 Likg x0.002 kg/d, _ 0.34
Oke ~ * myks-d
4.19 Hazard index = Sum of the hazard quotients:
Risk
Pg 45
54000 Like,
2 mg)Lof 1.1. Ltrichloroethane: RED = 0.035 mg/kg-d
Lax 2 mei.
= 0.04 mglkg-d
DE Kg:
ADD „ _0.04mp/ke-4 1 14
Ho RD 0.035 mg/kg-d bs
004 my/Lofterachlorethyene; RD = 0010mrka-d
1 LAx004 mL,
if (50 ks
9008
Ha RD 0010mgkg-d 0010 nee
mg/l. of 1,-dichlorethylene; RID = 0009 mg/kg
1 Lx 01 mg/l
vo = A Fog 0.002 on
RID" 0009 mglkg-d 0005
Hazard Index = 1.14 + 0.08 + 022 = 1.44, cause for concen.
4.20 1.0 gay of heptachlr leaking into a 30,000 m? pond: K=0.35lday:
a. Inputrate= Output ate + Decay ate=KCV (1.18)
1.0 g/day = 035/day x C(mg/L) x 103 g/mg x 30.000 m? x 103 Lim?
10
30,000:035
= 0.95x10"%mg/ 1.
b. 70-kg person, drinking 2 Liday for 5 years:
Risk = CDI x poteney
2L4 x 095k10*mg/Lx Sys _34
0 Kg x70 yes Cr
66x10 0.710%
4.21 0.03 mg of BaP per cigarette, 20 cigarettes/day for 40 years, potency 6.11 (me/kg-d)-!
20 cigld x 0.03 myeig x40 yrs
1 = 20 ig/d 0.03 maleig x40 VS 9.0049 mg -d
= 70 kgx 70 yrs Ye
Risk = CDIx potency = 0.0049 mgKg-4x6.11 (mgfkg-<)"= 003
Pg. 46
Lax 2 mei.
= 0.04 mglkg-d
DE Kg:
ADD „ _0.04mp/ke-4 1 14
Ho RD 0.035 mg/kg-d bs
004 my/Lofterachlorethyene; RD = 0010mrka-d
1 LAx004 mL,
if (50 ks
9008
Ha RD 0010mgkg-d 0010 nee
mg/l. of 1,-dichlorethylene; RID = 0009 mg/kg
1 Lx 01 mg/l
vo = A Fog 0.002 on
RID" 0009 mglkg-d 0005
Hazard Index = 1.14 + 0.08 + 022 = 1.44, cause for concen.
4.20 1.0 gay of heptachlr leaking into a 30,000 m? pond: K=0.35lday:
a. Inputrate= Output ate + Decay ate=KCV (1.18)
1.0 g/day = 035/day x C(mg/L) x 103 g/mg x 30.000 m? x 103 Lim?
10
30,000:035
= 0.95x10"%mg/ 1.
b. 70-kg person, drinking 2 Liday for 5 years:
Risk = CDI x poteney
2L4 x 095k10*mg/Lx Sys _34
0 Kg x70 yes Cr
66x10 0.710%
4.21 0.03 mg of BaP per cigarette, 20 cigarettes/day for 40 years, potency 6.11 (me/kg-d)-!
20 cigld x 0.03 myeig x40 yrs
1 = 20 ig/d 0.03 maleig x40 VS 9.0049 mg -d
= 70 kgx 70 yrs Ye
Risk = CDIx potency = 0.0049 mgKg-4x6.11 (mgfkg-<)"= 003
Pg. 46
4.22 Sidestream smoke, 6x10-4 mg/m3 while breathing 0.83 m3/hr, 106 risk, BaP potency
Sa anal
SAO mgeig m’ KOS MP ANGIE. quo"
E = 2800" N mg -d
U rm ON mE
Risk = CDIxPotency = 10%= 28x10“ Nimg/kg- dx ll
7 a N rd
10%
na
TEO Pr GAL
= 584 cigarettes
At eight cigarettes per day smoked in
EN
Seige
is poor fellow's presence:
days
4.23 Infilraion of 120 m3/hr, 0.1 mg BaPlcigarete, {cigarette per hour:
a. Find the steady-state concentration
Input rate = Output rate
1 cighrx0.1 mglig = 120 m7 x Cmg/n
C=0.1/120 = 0.00083 mans
b. living fora year witha smoker:
br
Doors 20 x LH Kg À x 265 ay
‘pt dy Zihr dy 10mg kg
sg FO Kg x 365 ye x70 ye minors te
Ris! 7106
CDI x Potency = 1.1x10-6 mg/kg-d x 6.11 (mg/kg-d)"
au Mange, un
cancer = 22X10 people x 6.6510 cancer person We _ 5 snceiye
—_— “Oyieime 2 Seancerly
b. benzene at 0.005 mg/L:
Pe. 47
Sa anal
SAO mgeig m’ KOS MP ANGIE. quo"
E = 2800" N mg -d
U rm ON mE
Risk = CDIxPotency = 10%= 28x10“ Nimg/kg- dx ll
7 a N rd
10%
na
TEO Pr GAL
= 584 cigarettes
At eight cigarettes per day smoked in
EN
Seige
is poor fellow's presence:
days
4.23 Infilraion of 120 m3/hr, 0.1 mg BaPlcigarete, {cigarette per hour:
a. Find the steady-state concentration
Input rate = Output rate
1 cighrx0.1 mglig = 120 m7 x Cmg/n
C=0.1/120 = 0.00083 mans
b. living fora year witha smoker:
br
Doors 20 x LH Kg À x 265 ay
‘pt dy Zihr dy 10mg kg
sg FO Kg x 365 ye x70 ye minors te
Ris! 7106
CDI x Potency = 1.1x10-6 mg/kg-d x 6.11 (mg/kg-d)"
au Mange, un
cancer = 22X10 people x 6.6510 cancer person We _ 5 snceiye
—_— “Oyieime 2 Seancerly
b. benzene at 0.005 mg/L:
Pe. 47
Risk = 240.005 mg L x 360 dlyrx30 yr, 2.9810
TO kg x 365 diyr x 70 yr mglke-d
“cancer = 260510" people x 1.7510 “cancer / person = fe
TOyrhifetime
75x10 <
©. arsenic at 0.05 mg/L:
2L/4 x 0.05 mgll x 360 dlyrx 30 yr. _1.75
70 kg x 365 dye x 70 yr md
260x10° people x 1.05x10"'cancer/ perso
Toyrlifeime
Risk 210510
ie
A cancer
4. carbon tetrachloride at 0.005 mg/L:
2LLa x 0.005 mg/L. x 360 diyr x 30 yr
"FO kg x 365 diye x 70 yr make -d
260x10" people x 7.9x10 “cancer /p if
TOyrlifeime
Risk = 9x0
& cancer =
+. vinyl chloride at 0.002 mg/L:
Risk = ZUAROMA mg/L x 360 dr x 30 yr, _ 23 _
TO Ke x 365 diye x 70 yr mgke-d
260x10* people x 5.6X10 ‘cancer / person = life
TOyrifeime
56x10"
A cancer
1. PCBs at 0.0005 mg/L:
21/4 x 0.0005 mg/L x 360 dlyrx30yr 7.7
TO kg x 365dlyr x70 yr Erz
260x10' people x 4.6x10"cancer/ person life
Toyrhifeime
Risk =. = 46x10
à cancer
4.25 Formaldehyde at 0 g/m} with potency 1.3x10°S cancer per ugin?
Risk
SOngin? x 13x}0-Seancer perygin’ = 65xt0-4
4.26 Groundwater with 10 ppb TCE vs. surface water with 5 ppb chloroform:
2L/d x 0.010 mg/L x 360 dlyrx30 yr | 1.Ix 1
O kg x 365 dlyr x70 ye ÊTRE
Pe. 48
Scancerÿr
fe 29 canceryr
=206 cancerlyr
72 cancerlyr
TO kg x 365 diyr x 70 yr mglke-d
“cancer = 260510" people x 1.7510 “cancer / person = fe
TOyrhifetime
75x10 <
©. arsenic at 0.05 mg/L:
2L/4 x 0.05 mgll x 360 dlyrx 30 yr. _1.75
70 kg x 365 dye x 70 yr md
260x10° people x 1.05x10"'cancer/ perso
Toyrlifeime
Risk 210510
ie
A cancer
4. carbon tetrachloride at 0.005 mg/L:
2LLa x 0.005 mg/L. x 360 diyr x 30 yr
"FO kg x 365 diye x 70 yr make -d
260x10" people x 7.9x10 “cancer /p if
TOyrlifeime
Risk = 9x0
& cancer =
+. vinyl chloride at 0.002 mg/L:
Risk = ZUAROMA mg/L x 360 dr x 30 yr, _ 23 _
TO Ke x 365 diye x 70 yr mgke-d
260x10* people x 5.6X10 ‘cancer / person = life
TOyrifeime
56x10"
A cancer
1. PCBs at 0.0005 mg/L:
21/4 x 0.0005 mg/L x 360 dlyrx30yr 7.7
TO kg x 365dlyr x70 yr Erz
260x10' people x 4.6x10"cancer/ person life
Toyrhifeime
Risk =. = 46x10
à cancer
4.25 Formaldehyde at 0 g/m} with potency 1.3x10°S cancer per ugin?
Risk
SOngin? x 13x}0-Seancer perygin’ = 65xt0-4
4.26 Groundwater with 10 ppb TCE vs. surface water with 5 ppb chloroform:
2L/d x 0.010 mg/L x 360 dlyrx30 yr | 1.Ix 1
O kg x 365 dlyr x70 ye ÊTRE
Pe. 48
Scancerÿr
fe 29 canceryr
=206 cancerlyr
72 cancerlyr
AL x 0.050 mg/L x360 dlyrx30yr 61x10
7x10
“FO kg x 365 diye x70 yr a 0
Chloroform risk =
Stick withthe groundwater.
4.27 70kg man, 0.1 myn? errchloroihlene, 1 mr, SiS Wink, $0 whye 30 ys,
0.9 absorption, potency 2x10°3 (mg/kg-d} risk
Img/m? x 1 m’/hr x Sheid x Sdlwk x SOwh/yr x 30 yr x
pe -20:10*a8
a TO kg x 365 diyr x 70 yr O10? mg! ke
Risk =CDI x poteney = 3.0x10'mg/kg=dx == = 6x10" forthe 70kg man
Risk for the 50 - kg woman = 6x10° x10
the 50-kg ET
4.28 Potency 03 (mg/kg-d)"!
ga
uns
mt
—_ “100 mi Ñ
Tae
time downstream = Om _, 4 1667days
T itr 247
L xe" .066mg/L
(066mg Lx AL x OU Jr
70kg x 365d/yr x 7Oyr 000796
col
Risk =0.000796 mg/kg -4 x 2440
030
mega
4.29 One-hitmodel:
Pld)=1
expl-(a0 + 4:d)] = 1 —exp[-(0.01209+0.001852xa)]
Multistage model:
PU =1 —expf as + qd + qe ++]
2 1- exp -{0.02077-+110IO % +1276m0 6 )]
Dose actual
multistage
Pg. 49
7x10
“FO kg x 365 diye x70 yr a 0
Chloroform risk =
Stick withthe groundwater.
4.27 70kg man, 0.1 myn? errchloroihlene, 1 mr, SiS Wink, $0 whye 30 ys,
0.9 absorption, potency 2x10°3 (mg/kg-d} risk
Img/m? x 1 m’/hr x Sheid x Sdlwk x SOwh/yr x 30 yr x
pe -20:10*a8
a TO kg x 365 diyr x 70 yr O10? mg! ke
Risk =CDI x poteney = 3.0x10'mg/kg=dx == = 6x10" forthe 70kg man
Risk for the 50 - kg woman = 6x10° x10
the 50-kg ET
4.28 Potency 03 (mg/kg-d)"!
ga
uns
mt
—_ “100 mi Ñ
Tae
time downstream = Om _, 4 1667days
T itr 247
L xe" .066mg/L
(066mg Lx AL x OU Jr
70kg x 365d/yr x 7Oyr 000796
col
Risk =0.000796 mg/kg -4 x 2440
030
mega
4.29 One-hitmodel:
Pld)=1
expl-(a0 + 4:d)] = 1 —exp[-(0.01209+0.001852xa)]
Multistage model:
PU =1 —expf as + qd + qe ++]
2 1- exp -{0.02077-+110IO % +1276m0 6 )]
Dose actual
multistage
Pg. 49
125 0.03 0216 0.044
250 023 0378 0215
500 0.88 0.608 0.888
ï ? 00138 00206
Problem 4.29 Model Predictions and Actual tumors
10.
08
T
| | much
|
os
04
Tumors/animal
02 |
0.0% al T
O 100 200 300 400 500 600
pom
4.30 10 million people, 10° risk,
cer ate 19510 people x 10 *ance person He
mes Foyer .
143 cancertyr
@10°5 risk, there would be 1.4 cancers, saving 143-1.4= 12.9 cancerlyr
Sliye- person x 10” people
129 cancers avoided
cost percancer
$O.77milion /cancer avoided
Pg. 4.10
250 023 0378 0215
500 0.88 0.608 0.888
ï ? 00138 00206
Problem 4.29 Model Predictions and Actual tumors
10.
08
T
| | much
|
os
04
Tumors/animal
02 |
0.0% al T
O 100 200 300 400 500 600
pom
4.30 10 million people, 10° risk,
cer ate 19510 people x 10 *ance person He
mes Foyer .
143 cancertyr
@10°5 risk, there would be 1.4 cancers, saving 143-1.4= 12.9 cancerlyr
Sliye- person x 10” people
129 cancers avoided
cost percancer
$O.77milion /cancer avoided
Pg. 4.10
SOLUTIONS FOR CHAPTER 5
5.1 In astandard BOD test:
a. orgerdto prevent reaction,
À. lito preven photos
e. tokeep final DO above zero,
8. toassure adequate microbial population
e: would take too long
5.2 BODS=200 mg/L; 90% efficient removal; initial DO=92 mg/L
pO,-DO,
F
=0.9)x200mg /L =20mg/L=
20mgL
EL 2036 (36% ofthe 300 mL is wastewater
Volume of treated wastewater =
d. 50% treated wastewater (P=0.50)
136 x 300mL = 108 mL.
DO, = DO, - Px BOD, = 9.2 -0.50x20
08mg/L
but DO cannot be negative, so DO = 0.
BOD, Z0mg/L
which corresponds to a volume of wastewater = 0,0087 x 300 mL = 2.6 mL (min)
To have final DO at least 2.0 mg/L means:
_DO,-DO, „80-2.0mg/L
E 006
which means max waste volume:
5.5 The BOD fortreated and untreated waste is
DO,-DO, _ 60-2008 L met
nated
u Bon: P 5/300mg/L.
Pg. 51
5.1 In astandard BOD test:
a. orgerdto prevent reaction,
À. lito preven photos
e. tokeep final DO above zero,
8. toassure adequate microbial population
e: would take too long
5.2 BODS=200 mg/L; 90% efficient removal; initial DO=92 mg/L
pO,-DO,
F
=0.9)x200mg /L =20mg/L=
20mgL
EL 2036 (36% ofthe 300 mL is wastewater
Volume of treated wastewater =
d. 50% treated wastewater (P=0.50)
136 x 300mL = 108 mL.
DO, = DO, - Px BOD, = 9.2 -0.50x20
08mg/L
but DO cannot be negative, so DO = 0.
BOD, Z0mg/L
which corresponds to a volume of wastewater = 0,0087 x 300 mL = 2.6 mL (min)
To have final DO at least 2.0 mg/L means:
_DO,-DO, „80-2.0mg/L
E 006
which means max waste volume:
5.5 The BOD fortreated and untreated waste is
DO,-DO, _ 60-2008 L met
nated
u Bon: P 5/300mg/L.
Pg. 51
‘Treated BOD , = PATO,
% BOD removal
From the graph
a. Ultimate BOD, Lo =40 mg/L.
b. BOD5=40- 15= 25 mg/L
e, Remaining BOD after 5 days = LS = 15 mg/L
BODS = 200 mg/L and Lo =300 mg/L
BOD, = La(1-e#)= 200mg/ L = 300mg/ Le)
200
uf BR) 022/60
„{1=10%) =200m8/L=300m8/L(1 10%)
200) 0.005,
1200) 0.0051,
om 00951 day
5.8 Given P= 1/3, DO¡=9.0 mg/L, DOS=40 mg/L, DO===2.0 mg/L:
% BOD removal
From the graph
a. Ultimate BOD, Lo =40 mg/L.
b. BOD5=40- 15= 25 mg/L
e, Remaining BOD after 5 days = LS = 15 mg/L
BODS = 200 mg/L and Lo =300 mg/L
BOD, = La(1-e#)= 200mg/ L = 300mg/ Le)
200
uf BR) 022/60
„{1=10%) =200m8/L=300m8/L(1 10%)
200) 0.005,
1200) 0.0051,
om 00951 day
5.8 Given P= 1/3, DO¡=9.0 mg/L, DOS=40 mg/L, DO===2.0 mg/L:
5.10
5.12
b. Sample 2at end of test:
D0,=
P00, =92-Lsai0=-a8n/L
but DO cant be negative, so DOf
Do,
BOD,
= Omg.
2
©. DOF>20, so: PS =005
140
max Vol treated wastewater = 0.05 x 300 mL = 15 mL
Seeded waste: 1:30, DO} =9.2, DO[=2.0 mg/L
Blank: Bj=92, Bf=8.0 mg/L, Using (58) gives,
_(00,~00,)-(8,-B.\1~P)_92-20-02-80f1- Yo) y,
7 130
Using (58) gives,
gop,» (29.-09)-(B-BXI-"
835-240 (875-891
- 110
BOD5=150mg/L at 20°C, k=0.23/day:
2. Tofind ultimate BOD, rearrange (5.12):
150mg/L
19.5mg/L
1. To find kat 15°C, use (5.15):
kr
287%; kg m028x (1.08799 0.183 day
Pg. 53
5.12
b. Sample 2at end of test:
D0,=
P00, =92-Lsai0=-a8n/L
but DO cant be negative, so DOf
Do,
BOD,
= Omg.
2
©. DOF>20, so: PS =005
140
max Vol treated wastewater = 0.05 x 300 mL = 15 mL
Seeded waste: 1:30, DO} =9.2, DO[=2.0 mg/L
Blank: Bj=92, Bf=8.0 mg/L, Using (58) gives,
_(00,~00,)-(8,-B.\1~P)_92-20-02-80f1- Yo) y,
7 130
Using (58) gives,
gop,» (29.-09)-(B-BXI-"
835-240 (875-891
- 110
BOD5=150mg/L at 20°C, k=0.23/day:
2. Tofind ultimate BOD, rearrange (5.12):
150mg/L
19.5mg/L
1. To find kat 15°C, use (5.15):
kr
287%; kg m028x (1.08799 0.183 day
Pg. 53
Le) =2195f1 °°") = 131 Smg/L
5.13 BOD5at200C=210 mg/L and Lo =350 mglL, find BODS at 25°C:
Lele
(850-2102 350.0“
1 [140
= big 0) 20.185216
Kır 5" (355) Y
To find kar 25°C, use (5.15),
kg, 1832 x (1.047)%9=0.231/day
ke = ka =
=) =350(1 2%) =2395m8/L
‘Then, BOD,(25°C)= Lol
5.14 Plots of BODts1 vs BOD along with the line representing BOD41=BOD: are shown.
Finding Lo has considerable room for error.
Waste 1
MET ET
350 +
300 :
250 E
200 :
+
C0
150
100 {a
so E
o
T Ht
+ +
= H
O 50 100 150 200 250 300 350 400
BODE Lo=270m9/L
T
HE
+t
Pg. 54
5.13 BOD5at200C=210 mg/L and Lo =350 mglL, find BODS at 25°C:
Lele
(850-2102 350.0“
1 [140
= big 0) 20.185216
Kır 5" (355) Y
To find kar 25°C, use (5.15),
kg, 1832 x (1.047)%9=0.231/day
ke = ka =
=) =350(1 2%) =2395m8/L
‘Then, BOD,(25°C)= Lol
5.14 Plots of BODts1 vs BOD along with the line representing BOD41=BOD: are shown.
Finding Lo has considerable room for error.
Waste 1
MET ET
350 +
300 :
250 E
200 :
+
C0
150
100 {a
so E
o
T Ht
+ +
= H
O 50 100 150 200 250 300 350 400
BODE Lo=270m9/L
T
HE
+t
Pg. 54
opts
5.15
Waste 2
300 — y
250 er y
200
150 1
400 1
|
50 —
TNT
© so 10 150 200 250 200
Bove L-220mgL
Waste 3
= | dB
Pi EEE
ET
+ LA
200 II 4
z +| t
2 150 >
E 10 —
i
TZ 1 i
T HOT
3o 100 150 200 20 300
SD Lo-tsömg/t
Show BODy+1 = a BOD; + bibatis, tha tis linear
BOD,,, = L,(I-e™* se)
now add and subtracte
BOD,,, =L,li-etettet-et
BoD, = et L{i=e*)+L (ie
BOD,,,=aBOD, +b where
Pe. 55
5.15
Waste 2
300 — y
250 er y
200
150 1
400 1
|
50 —
TNT
© so 10 150 200 250 200
Bove L-220mgL
Waste 3
= | dB
Pi EEE
ET
+ LA
200 II 4
z +| t
2 150 >
E 10 —
i
TZ 1 i
T HOT
3o 100 150 200 20 300
SD Lo-tsömg/t
Show BODy+1 = a BOD; + bibatis, tha tis linear
BOD,,, = L,(I-e™* se)
now add and subtracte
BOD,,, =L,li-etettet-et
BoD, = et L{i=e*)+L (ie
BOD,,,=aBOD, +b where
Pe. 55
5.16 BODS=180mg/L, k=0.22/day, TKN=30mg/L:
BOD, 180
Te) Fe
b. NBOD =4.57XTKN = 4,57x30mg/L=137mg/L
a CBOD=L, =
©. BODremaining = (270 + 137)- 180 =227 mg/L.
5.17 CóH1506N + 602 > 6CO2 + 6 H20 + NH3
mol wt algae = 6x12+ 15 + 6x16 + 14= 197 g/mol
2. 6 moles of O2 (6x2x16=192 g) oxidizes 1 mole (197 g) of algae
10 mgalgie „19230,
carbonaceous THOÏ Te
75 mg/L
b. Nitrogenous portion:
NH3 +202 > NO3" +H* + H20
1 mole NH3 (178) needs 2 moles O (€
JOmpalgre , meh, G48 Os asm)
NBOD =
I97mgalgae "17mg NA,
‘Total theoretical oxygen demand = 9,75 à 3.25 = 13.0 mg/L.
5.18 2 CHANH2)COOH +3 02 — 4C02 + 2 H20 + 2 NH3
mol wt glycine = 2x12 + Sel + Ix14+2x16=75 g/mol
3molO, _,_32g0,/mol_ 200mg glycine
8 CBOD = gaine * 75g glycine/mol L
=128mg/L
b. Nitrogenous portion: NH3 +202 > NO3 + Ht + H20
Neon = 21010, , 2m01 NH, , 32g/molO, 200mg BRE ne,
Imol NH, “Imol glycine * 75g/mol glycine _L
©. Total theoretical oxygen demand = 128 + 171 =299 mg/l.
5.19 C8H1203N2 +802 => 8 CO2 + 3 H20 +2 NH3
mol wt casein = 8x12 + 12x1 +3x16 + 2x14= 184 g/mol
Pg. 56
BOD, 180
Te) Fe
b. NBOD =4.57XTKN = 4,57x30mg/L=137mg/L
a CBOD=L, =
©. BODremaining = (270 + 137)- 180 =227 mg/L.
5.17 CóH1506N + 602 > 6CO2 + 6 H20 + NH3
mol wt algae = 6x12+ 15 + 6x16 + 14= 197 g/mol
2. 6 moles of O2 (6x2x16=192 g) oxidizes 1 mole (197 g) of algae
10 mgalgie „19230,
carbonaceous THOÏ Te
75 mg/L
b. Nitrogenous portion:
NH3 +202 > NO3" +H* + H20
1 mole NH3 (178) needs 2 moles O (€
JOmpalgre , meh, G48 Os asm)
NBOD =
I97mgalgae "17mg NA,
‘Total theoretical oxygen demand = 9,75 à 3.25 = 13.0 mg/L.
5.18 2 CHANH2)COOH +3 02 — 4C02 + 2 H20 + 2 NH3
mol wt glycine = 2x12 + Sel + Ix14+2x16=75 g/mol
3molO, _,_32g0,/mol_ 200mg glycine
8 CBOD = gaine * 75g glycine/mol L
=128mg/L
b. Nitrogenous portion: NH3 +202 > NO3 + Ht + H20
Neon = 21010, , 2m01 NH, , 32g/molO, 200mg BRE ne,
Imol NH, “Imol glycine * 75g/mol glycine _L
©. Total theoretical oxygen demand = 128 + 171 =299 mg/l.
5.19 C8H1203N2 +802 => 8 CO2 + 3 H20 +2 NH3
mol wt casein = 8x12 + 12x1 +3x16 + 2x14= 184 g/mol
Pg. 56
coop =—2m1O,_,3280,/mel_, 200mg enn
mol easein * IS4gcascinimal
b. Nitrogenous portion: NH3+ 202 > NOy + H* + H20
NBOD = x >
{mol NH, * Imolcascin * 184g/molcascin
e. Total BOD = 278 + 139 =417 mg/L.
4. BOD,= CBOD(I~¢
= 278{1- ee) D
5.20 CSH7O2N + 502 > 5CO2 + 2 H20 + NH3
mol wt eelis = Sx12+7x1 + 2x16 + 14= 113 g/mol
SmolO, , 3280,/mol
a „Sm, , 32805 mA, 12 cats «141650,
BoD mol cells 113gcells/mol © 121680,
b. Nitrogenous portion: NH3 + 202 > NO3" + H* + H20
2molO, Imol NH, 32g/mol 0,
me mol NH, * Tmolcelts * 113g/mol cells
e. Total BOD = 1.416 + 0.566 = 1.98 g 02 per g cells
5.21
d= 0.22/6
unam SIR waa mg
Gomas N tm
a. just downstream:
yw Sabet QE In x Omg + VOA SM song,
QQ, 10+ ms
b. at 10,000 m downstream:
10,000m he, day
Tim’ /s, 36005 * 24he
yaa
=0.578days
TE mg/L
Pg. 57
Eng IL
2molO, , 2mol NH, , 32g/molO, „200mgeasein
L
Parma, y yg cells =
mol easein * IS4gcascinimal
b. Nitrogenous portion: NH3+ 202 > NOy + H* + H20
NBOD = x >
{mol NH, * Imolcascin * 184g/molcascin
e. Total BOD = 278 + 139 =417 mg/L.
4. BOD,= CBOD(I~¢
= 278{1- ee) D
5.20 CSH7O2N + 502 > 5CO2 + 2 H20 + NH3
mol wt eelis = Sx12+7x1 + 2x16 + 14= 113 g/mol
SmolO, , 3280,/mol
a „Sm, , 32805 mA, 12 cats «141650,
BoD mol cells 113gcells/mol © 121680,
b. Nitrogenous portion: NH3 + 202 > NO3" + H* + H20
2molO, Imol NH, 32g/mol 0,
me mol NH, * Tmolcelts * 113g/mol cells
e. Total BOD = 1.416 + 0.566 = 1.98 g 02 per g cells
5.21
d= 0.22/6
unam SIR waa mg
Gomas N tm
a. just downstream:
yw Sabet QE In x Omg + VOA SM song,
QQ, 10+ ms
b. at 10,000 m downstream:
10,000m he, day
Tim’ /s, 36005 * 24he
yaa
=0.578days
TE mg/L
Pg. 57
Eng IL
2molO, , 2mol NH, , 32g/molO, „200mgeasein
L
Parma, y yg cells =
OF = 10 m/s =1m3/ DOsat10.08mg/L
Bors malt AN) Oye Amat IS,
a. just downstream:
40mg/L x 1m’/s+8mg/Lx 10m"/s
nO 1410 m/s
=74mg/L
From Table 5.11, DOsa
0.08 mg/L.
Initial deficit Do = DOsa - DO = 10.08 -7.64=2.44 mg/L.
5.23 Notice when D, =
then D = ute (0% 071) thats the deficit is proportional
tothe initial BOD. Doubling Lo will therefore double the deficit at every point
Before: maximum deficit Dmax = DOsat- DOmin = 10.0-6.0
mel
After: maximum deficit Dmax=2x 4mg/L = 8 mg/L.
now DOmin = DOsst - Dmax = 10.0- 8.0 = 2.0 mg/L.
5.24
Do (man)
Asin Prob. 5.23, deficit is proportional to BOD:
original Dmax = 10.0 - 3.0 = 7.0 mg/L.
desired Dmax = 10.0 - 50 = 5.0 mg/L.
SOmg/L _ 971
7Omg/L”
‘Therefore, need to remove 1.0- 0,71 20.29 = 29% of iit
BoD.
Bors malt AN) Oye Amat IS,
a. just downstream:
40mg/L x 1m’/s+8mg/Lx 10m"/s
nO 1410 m/s
=74mg/L
From Table 5.11, DOsa
0.08 mg/L.
Initial deficit Do = DOsa - DO = 10.08 -7.64=2.44 mg/L.
5.23 Notice when D, =
then D = ute (0% 071) thats the deficit is proportional
tothe initial BOD. Doubling Lo will therefore double the deficit at every point
Before: maximum deficit Dmax = DOsat- DOmin = 10.0-6.0
mel
After: maximum deficit Dmax=2x 4mg/L = 8 mg/L.
now DOmin = DOsst - Dmax = 10.0- 8.0 = 2.0 mg/L.
5.24
Do (man)
Asin Prob. 5.23, deficit is proportional to BOD:
original Dmax = 10.0 - 3.0 = 7.0 mg/L.
desired Dmax = 10.0 - 50 = 5.0 mg/L.
SOmg/L _ 971
7Omg/L”
‘Therefore, need to remove 1.0- 0,71 20.29 = 29% of iit
BoD.
5.25
mel
200
reraining
5.26 a. Long after teatment plant breaks down, deficit is doubled since the BOD is
doubled.
before breakdown
ong after breakdown
0123456 y ——
b, Only 4 days after the breakdown, the first 4-days beond the outfall have been
affected, beyond that, the DO is same as before the breakdown:
before breakdown
A
> a days after brescconmn
0123456 ays)
DO (mg)
inimum DO of 3.0 mg/L means the maximum deficit (before fixing it) is
"mg/L. For healthy conditions, we want DO pin to be S mg/L,
0-5 =SmgiL.
5.27 a. The
Dimas = 10-
so that means we want Dmax(new)
Dec . SOmg/L 97]
Dane 100-3.0m8/L
Pg. 59
mel
200
reraining
5.26 a. Long after teatment plant breaks down, deficit is doubled since the BOD is
doubled.
before breakdown
ong after breakdown
0123456 y ——
b, Only 4 days after the breakdown, the first 4-days beond the outfall have been
affected, beyond that, the DO is same as before the breakdown:
before breakdown
A
> a days after brescconmn
0123456 ays)
DO (mg)
inimum DO of 3.0 mg/L means the maximum deficit (before fixing it) is
"mg/L. For healthy conditions, we want DO pin to be S mg/L,
0-5 =SmgiL.
5.27 a. The
Dimas = 10-
so that means we want Dmax(new)
Dec . SOmg/L 97]
Dane 100-3.0m8/L
Pg. 59
so, we need to remove 29% of the BOD. Since a primary treatment plant
removes about 35% of the BOD (Chapter 6) it should do the job.
b. Using (531) we can find the critical time and distance downstream:
fte 1/08) 931 days
le e
distane
60 milday x 231 days = 138.6 miles
e, What ultimate BOD to assure 5 mg/L (Eqn. 527):
kL
Ka (ee)
ka
A
wee est) 20.157,
4
139 mies amie) ——
1 (23)
a =
550203018103)" 1 és
48.0 milefday x 1.83 days =879 miles
».
2314 x 00 mL eus
(09 - 03a
DOrnin = DOsat - Dax
Pg. 510
removes about 35% of the BOD (Chapter 6) it should do the job.
b. Using (531) we can find the critical time and distance downstream:
fte 1/08) 931 days
le e
distane
60 milday x 231 days = 138.6 miles
e, What ultimate BOD to assure 5 mg/L (Eqn. 527):
kL
Ka (ee)
ka
A
wee est) 20.157,
4
139 mies amie) ——
1 (23)
a =
550203018103)" 1 és
48.0 milefday x 1.83 days =879 miles
».
2314 x 00 mL eus
(09 - 03a
DOrnin = DOsat - Dax
Pg. 510
052
therefore, need to remove 1.0- 0.52 = 48% of the BOD.
5.29 Now the river has an initial deficit that is not 0, so we can’t use the simple
proportionality between Lo and Dmax
(
.—_ KT, _ Dotk, al 1 (09[,_209-03) en
eel som oe Ja
critical distance = 48 mifday x 1.69 days = 81.2 miles
RER
v.D, rad
Ener
0.3/4 x 500 mglL {our ¿00069 4 > 99-0006,
= PEDO mE e = 713
a ) + 2.00" 10.03 mg,
DOmin = 10.0- 10.03<0 anaerobic, = DOmin =0
5.30
DOr = 7.6 mg/L. u
236 mg/L rer
EE = Sn
PR
0.518
u=12f/s DOsat=8.5 mg/L ke =0.76/d
a. initial conditions:
Tefs x 1SmglL + 250efs x 7.6mg/L.
1) 37 + 250cfs somal
Inia deficit = Dy =8.Smg/L~685mg/L= L.65mg/L
Telex mg + 250efs x3.6mg/l-_ 6
hi a Ly = Es 28m + 25061 Img _ 75
iia BOD = Ly meee 6:75mg/L
| ner TK 2)
Pg SAL
therefore, need to remove 1.0- 0.52 = 48% of the BOD.
5.29 Now the river has an initial deficit that is not 0, so we can’t use the simple
proportionality between Lo and Dmax
(
.—_ KT, _ Dotk, al 1 (09[,_209-03) en
eel som oe Ja
critical distance = 48 mifday x 1.69 days = 81.2 miles
RER
v.D, rad
Ener
0.3/4 x 500 mglL {our ¿00069 4 > 99-0006,
= PEDO mE e = 713
a ) + 2.00" 10.03 mg,
DOmin = 10.0- 10.03<0 anaerobic, = DOmin =0
5.30
DOr = 7.6 mg/L. u
236 mg/L rer
EE = Sn
PR
0.518
u=12f/s DOsat=8.5 mg/L ke =0.76/d
a. initial conditions:
Tefs x 1SmglL + 250efs x 7.6mg/L.
1) 37 + 250cfs somal
Inia deficit = Dy =8.Smg/L~685mg/L= L.65mg/L
Telex mg + 250efs x3.6mg/l-_ 6
hi a Ly = Es 28m + 25061 Img _ 75
iia BOD = Ly meee 6:75mg/L
| ner TK 2)
Pg SAL
5.31
1 fe 164076-060))). 95 Jays
Gre-cena"(oal 7 seas 3)
a
critical distance
abc frac satay 07m
©. minimum DO:
de fo eee"
Pane tle ede
Ud x 6.75 MB nos ¿270009 y gg
= SIRT EE (ates 1701) sten 285 mg/L
DOmin = DOsat - Dmax =8.5-285=56 mg/L.
4. 10 miles downstream:
e On x S28OfUmi
Ts x 3600%/he x 24he/d
Jen
= 0.Sléays
978035 y SE mg/L
59 mg/L
00" 6.0 mn
O REA
a DS) Meee
PA
um 0.65 m/s DOsat = 8.0 mg/L Kt=0.37/8 kd= 0.2076
Initial conditions:
m/s x LOmg/L + Om x
po - Im /5x LOmg/L + Om sx 6mg/L
47508
03 + 09m /s APE
Ina def Do =8.0-475 2325 me.
" fax da
iii BOD = 1, „ SANGAMON TOME a
Pg. 512
1 fe 164076-060))). 95 Jays
Gre-cena"(oal 7 seas 3)
a
critical distance
abc frac satay 07m
©. minimum DO:
de fo eee"
Pane tle ede
Ud x 6.75 MB nos ¿270009 y gg
= SIRT EE (ates 1701) sten 285 mg/L
DOmin = DOsat - Dmax =8.5-285=56 mg/L.
4. 10 miles downstream:
e On x S28OfUmi
Ts x 3600%/he x 24he/d
Jen
= 0.Sléays
978035 y SE mg/L
59 mg/L
00" 6.0 mn
O REA
a DS) Meee
PA
um 0.65 m/s DOsat = 8.0 mg/L Kt=0.37/8 kd= 0.2076
Initial conditions:
m/s x LOmg/L + Om x
po - Im /5x LOmg/L + Om sx 6mg/L
47508
03 + 09m /s APE
Ina def Do =8.0-475 2325 me.
" fax da
iii BOD = 1, „ SANGAMON TOME a
Pg. 512
5.32
Critical time:
Maximum deficit
Dae la e
A
De
= 9206x685 mg {ar
= 7037-0204
DOmin = 80-33 =47 mg/l.
=
03 ms N te= 109 mg
DOsat (300€) = 7.6 mg/L. Don 7.6-6.3= 1.3 mg/L
Need to adjust for 30°C temperature (instead of 20°C):
from (5.19,
K,(0°C) =, (20°C}(.047°°29 = 0201 4 (1.047) *=03171day
from page 201,
(30°C) = k,(20°C 1.024)?" = 04t/a x (1.024)? =0:520/day
critical ime, t, Dicta]
Ke
0520], _130.520 cau]
10 2521, 130320030] )_ 2 05 days
(rl 03109 sen
ei ene = 4, = 038 00 a2 08
050m
maximum deficit,
Ei
min (et ee De
a
kun na) aay ate
an le 008204) 41368 mg/L
Pe. 513
Armes) 3.256% 03.3 mg/L
Critical time:
Maximum deficit
Dae la e
A
De
= 9206x685 mg {ar
= 7037-0204
DOmin = 80-33 =47 mg/l.
=
03 ms N te= 109 mg
DOsat (300€) = 7.6 mg/L. Don 7.6-6.3= 1.3 mg/L
Need to adjust for 30°C temperature (instead of 20°C):
from (5.19,
K,(0°C) =, (20°C}(.047°°29 = 0201 4 (1.047) *=03171day
from page 201,
(30°C) = k,(20°C 1.024)?" = 04t/a x (1.024)? =0:520/day
critical ime, t, Dicta]
Ke
0520], _130.520 cau]
10 2521, 130320030] )_ 2 05 days
(rl 03109 sen
ei ene = 4, = 038 00 a2 08
050m
maximum deficit,
Ei
min (et ee De
a
kun na) aay ate
an le 008204) 41368 mg/L
Pe. 513
Armes) 3.256% 03.3 mg/L
5.33
5.34
DO min =DOsa1(300C) -Dmax =7.6-3.5=4.1 mg/L.
(for comparison, at 200C it was 6.0 mg/L)
14-0208 2 to=20m0t
coco N, 002 60m.
PR
am um
a. Finding kr using (5.
ea 39025
Eu
” (El
2 1, (037s), 300575 0201]
a ra re
79 days
€: minimum DO,
Jl lo ep
De er
020120 {armure 45 gg
PRE ern 4300
DOmin = 9.0 - 6.1 =2.9 mg/L.
dl. What percent removal needed to get DOmin = 5.0 mg/L
D
D
o
0.66, so must remove 1.00 - 0.66 = 03:
Sl
‘Algae represented by C1OSH2630110N16P.
mol wt 10612 +263x1 + 110x16 + 16x14 + 31 =3550 g/mol
Pg. 5.14
5.34
DO min =DOsa1(300C) -Dmax =7.6-3.5=4.1 mg/L.
(for comparison, at 200C it was 6.0 mg/L)
14-0208 2 to=20m0t
coco N, 002 60m.
PR
am um
a. Finding kr using (5.
ea 39025
Eu
” (El
2 1, (037s), 300575 0201]
a ra re
79 days
€: minimum DO,
Jl lo ep
De er
020120 {armure 45 gg
PRE ern 4300
DOmin = 9.0 - 6.1 =2.9 mg/L.
dl. What percent removal needed to get DOmin = 5.0 mg/L
D
D
o
0.66, so must remove 1.00 - 0.66 = 03:
Sl
‘Algae represented by C1OSH2630110N16P.
mol wt 10612 +263x1 + 110x16 + 16x14 + 31 =3550 g/mol
Pg. 5.14
106x12
3550
H=28 =0.0741=74.1m8
35%
= 03583 =3583m8
= ISO 9. 4958= 495.8m8
a pert gotas 27 3550
14
2198 0.0631 = 63.1mg
n= 168 0061= 63.1mg
p= 21 -0.087=87ma
3550
b. with 0.10 mg N and 0.04 mg of P available,
Nallows: pm y lO0Ompalgar 1 6mg/ Lalgae
6.1 man
allows; 04 me? | I00Omealgae _ à Ge Latgae
Pal = = 46mg/ Lai
So, Nis limiting nutrient
+. mass of algae can be 1.6 mg algae/L,
de ke by 50"
et tar
5.35 Given 60mg N and 10mg P per 1000 mg algse. At 0.12 mgN/L and 0.03 mgP/L,
available:
0.12 mgN , 1000mgalgee
= 2.0mg al
= an 2Omgalge/L
0.03 mgP | 1000mgalgac.
10.0 maP
Omg algae /L.
Nitrogenis the limiting nutrient.
a. To cone alga production o 1.0 mg/L means N needs to be reduce to
comes Se eduction o
b. To contro algal production by reducing P requires 0.01 mgPIL, a 66.6% reduction.
Pg. 515
3550
H=28 =0.0741=74.1m8
35%
= 03583 =3583m8
= ISO 9. 4958= 495.8m8
a pert gotas 27 3550
14
2198 0.0631 = 63.1mg
n= 168 0061= 63.1mg
p= 21 -0.087=87ma
3550
b. with 0.10 mg N and 0.04 mg of P available,
Nallows: pm y lO0Ompalgar 1 6mg/ Lalgae
6.1 man
allows; 04 me? | I00Omealgae _ à Ge Latgae
Pal = = 46mg/ Lai
So, Nis limiting nutrient
+. mass of algae can be 1.6 mg algae/L,
de ke by 50"
et tar
5.35 Given 60mg N and 10mg P per 1000 mg algse. At 0.12 mgN/L and 0.03 mgP/L,
available:
0.12 mgN , 1000mgalgee
= 2.0mg al
= an 2Omgalge/L
0.03 mgP | 1000mgalgac.
10.0 maP
Omg algae /L.
Nitrogenis the limiting nutrient.
a. To cone alga production o 1.0 mg/L means N needs to be reduce to
comes Se eduction o
b. To contro algal production by reducing P requires 0.01 mgPIL, a 66.6% reduction.
Pg. 515
5.36
109/m p
1001108 m?
from (535), the phosphorus input allows a concent
of phosphorus of:
4m? /s x 10gP/m?
CAES
an 0.0778! m?
0.077m8/L
Nom /yrx 100k 10m
20.4m'/s +
3
ax
To keep phosphorus at less than 0.010 mg/L,
Source want _Cwant _0.010mg/L
Soureehave ~ Chave " OO7Img/L ©
Need to reduce the phosphorus input by I - 0.1;
void volume _ (50~35)em?
Totalvolume Som?
1008.
End
5.37 a. porosiy
bo solids density =. 2,86g/em?
5.38 Gravel aquifer, Table 5.12, porosity = 25%, specific yield = 22%:
.22 x 10,000 m2 x 1.0 m = 2200 m3
yield = specific yield x volume
200? 2 _
Porositysvolume ~ 0.25x10,000m’xim
fraction removed
Pg. 5.16
109/m p
1001108 m?
from (535), the phosphorus input allows a concent
of phosphorus of:
4m? /s x 10gP/m?
CAES
an 0.0778! m?
0.077m8/L
Nom /yrx 100k 10m
20.4m'/s +
3
ax
To keep phosphorus at less than 0.010 mg/L,
Source want _Cwant _0.010mg/L
Soureehave ~ Chave " OO7Img/L ©
Need to reduce the phosphorus input by I - 0.1;
void volume _ (50~35)em?
Totalvolume Som?
1008.
End
5.37 a. porosiy
bo solids density =. 2,86g/em?
5.38 Gravel aquifer, Table 5.12, porosity = 25%, specific yield = 22%:
.22 x 10,000 m2 x 1.0 m = 2200 m3
yield = specific yield x volume
200? 2 _
Porositysvolume ~ 0.25x10,000m’xim
fraction removed
Pg. 5.16
5.39
K3 = 100.3m
2, use as datum
(0.2m above datum
Om above datum
=0.1m above datum
34 (109,100)
tom
(0.1m)
CRE
(0.2m) (0.0m)
sta toun an hea ism: Dep an at
bea ey
5.41 From Prob. 539,
grad = 0.00115, hydraulic conductivity K = 1000m/, porosity
000m/ dx 0.0013:
ah
a. Darey velocity = v= KT 115 mida)
y velocity E y
Pg. 5.17
K3 = 100.3m
2, use as datum
(0.2m above datum
Om above datum
=0.1m above datum
34 (109,100)
tom
(0.1m)
CRE
(0.2m) (0.0m)
sta toun an hea ism: Dep an at
bea ey
5.41 From Prob. 539,
grad = 0.00115, hydraulic conductivity K = 1000m/, porosity
000m/ dx 0.0013:
ah
a. Darey velocity = v= KT 115 mida)
y velocity E y
Pg. 5.17
pm Darcy veloity Listin sud
. avg. linear velocit
ba avg. li iy: a a
300m
A
distance,d __300sin6o”
¿Om day
‘without accounting for retardation, {=
‘vg. velocity Y
with retardation factor of 2: t=R x 52422 x 52 d = 104 days
75
Nom
gm
= KAŸ 702 x(10m x 750m)
Q= KAT = 7.02 x(10mx 750m)
262.5m day
1000m
5.43 v'= LOnv/day, grad = 0.0005, porosity n = 020
Darey velocity = v= Kx grad
y _Yn_1Om/day £0.20
ad grad 0.0005
5.44
Pg. 518
. avg. linear velocit
ba avg. li iy: a a
300m
A
distance,d __300sin6o”
¿Om day
‘without accounting for retardation, {=
‘vg. velocity Y
with retardation factor of 2: t=R x 52422 x 52 d = 104 days
75
Nom
gm
= KAŸ 702 x(10m x 750m)
Q= KAT = 7.02 x(10mx 750m)
262.5m day
1000m
5.43 v'= LOnv/day, grad = 0.0005, porosity n = 020
Darey velocity = v= Kx grad
y _Yn_1Om/day £0.20
ad grad 0.0005
5.44
Pg. 518
de oh
Q= KATE = Korb
Peer
5.45
Q=1000m3/d
01m
hom
r
| @un() „100m saa(29% )
RS ee D
5.46
5000m3/d
Pg. 519
Q= KATE = Korb
Peer
5.45
Q=1000m3/d
01m
hom
r
| @un() „100m saa(29% )
RS ee D
5.46
5000m3/d
Pg. 519
Q UN (557
Om) mam)
o ZRH)
jos» aronjras ao
Qe ABR
EL
a QED.
5.48 From Prob. 546: B=30m, Q=5000m3/d, R=15m, rw=0.2m, n=0.30.
x030x 30m(15* -02*)m*
‘stagnation point. o ‘ ‘expanded view
an
from expanded view,
Pg. 520
Om) mam)
o ZRH)
jos» aronjras ao
Qe ABR
EL
a QED.
5.48 From Prob. 546: B=30m, Q=5000m3/d, R=15m, rw=0.2m, n=0.30.
x030x 30m(15* -02*)m*
‘stagnation point. o ‘ ‘expanded view
an
from expanded view,
Pg. 520
AE
d- tan 8 = öforsmall values of 6
Q (3)
Therefor, forsmall 6, y =p
Q thatis, x = - Q
zn OEP
5.50 0.1m3-TCE plume, B=10m, 2000m x250m , grad=0.001, K=0.001mis, h=04
a. Canit dissolve? Using the specific gravity given in Table 5.15, the concentration C
of TCE in the plume (iit eou all dissolve) would be
cu film’ x AT Kgl 1OL/ mx 10'm8/Kg _ 147310" ng
= "3000 x 250m x 10m x 040% OTL? 2x10"
Since the solubiliy (Table 5.15) is 1100 mg/L, YES it could all dissolve.
b. Try a single well:
‚Om /s x 0.001=1x10"'m/s
Q= 250 x 2Bv = 250m x 2 x 10m x 1xkO “m/s = 0.005m? /s
this exceeds the maximum pumping rate, which is given as 0.003 m/s,
Pg. 521
d- tan 8 = öforsmall values of 6
Q (3)
Therefor, forsmall 6, y =p
Q thatis, x = - Q
zn OEP
5.50 0.1m3-TCE plume, B=10m, 2000m x250m , grad=0.001, K=0.001mis, h=04
a. Canit dissolve? Using the specific gravity given in Table 5.15, the concentration C
of TCE in the plume (iit eou all dissolve) would be
cu film’ x AT Kgl 1OL/ mx 10'm8/Kg _ 147310" ng
= "3000 x 250m x 10m x 040% OTL? 2x10"
Since the solubiliy (Table 5.15) is 1100 mg/L, YES it could all dissolve.
b. Try a single well:
‚Om /s x 0.001=1x10"'m/s
Q= 250 x 2Bv = 250m x 2 x 10m x 1xkO “m/s = 0.005m? /s
this exceeds the maximum pumping rate, which is given as 0.003 m/s,
Pg. 521
Q=250mx 10m x Ix1O'm /s= 0.0025m? /s< 0,003m’ Is
80,2 wells will work.
©. spacing for minimum pumping rate:
Q___ 00025m/s
BBV 7x I0mx 10 m/s
spacing = 79.6 =80m apart
5.51. Show width of capture zone for n wells is nQU2B4),
y
7 1 %
Yo G?
12
on
mo reihe)
‘As shown in the figure, when x = 0, call the y value of the capture zone curve y90.
For x = 0, and y = y90, and n wells, all nof the 9; are 7/2, so
E a-ak
“The width of the capture zone along the y-axisis just2x yoo = Qn/(2Bv) QED.
Pg. 522
80,2 wells will work.
©. spacing for minimum pumping rate:
Q___ 00025m/s
BBV 7x I0mx 10 m/s
spacing = 79.6 =80m apart
5.51. Show width of capture zone for n wells is nQU2B4),
y
7 1 %
Yo G?
12
on
mo reihe)
‘As shown in the figure, when x = 0, call the y value of the capture zone curve y90.
For x = 0, and y = y90, and n wells, all nof the 9; are 7/2, so
E a-ak
“The width of the capture zone along the y-axisis just2x yoo = Qn/(2Bv) QED.
Pg. 522
5.52 Be20m,h = 030, K= 10-4m/s, grad = 0.0015
4-22
zur) Br
E x 20m x 1x10'm/sx 00015 x(10%4)m = 0.0004"
5.53 Im? of a sand-gravel aquifer, 201. of tetrachloroethylene, 20% of solubility,
a. From Table 5.12 the porosity is estimated at 0.20, from Table 5.15 the solubility is
given as 150 mg/L. Since the amount dissolved is only 20% ofits solubility:
20x 150 mg/l. x 1 m3 x 1034/3 x 0.20 = 6000 mg = 68
dissolved pere
b. Tofindreminingteachlroetylen, us th sp. gr. from Table 5.15 0F 1.63,
total mass=201 x 1.69 kyLx IB kg = 326003
undissolved mass = 32,600 632,594 g (almost ll not dissolved)
€: gradient = 0.001, porosity = 020,
average linear velocity = 2012 = SOME 3x 0.001 > 95 019. 1m/day
K(B4.) _ 410m/d x 0.001
7 020
a. using 1 m? cross section for oureubie meter of aquifer gives, andthe 6 gm?
Éontamhant concentration gives,
aquifer flow rate = 2.05 m/d x 1 m? = 2.05 mid
contaminant removal rate = 2.05 m3/d x 6 g/m? = 123 g/day
Lo remove the total 32,600 g would ta
pa 32,6008
123 /day x 3653ayiyr
=73 years
Pg 523
4-22
zur) Br
E x 20m x 1x10'm/sx 00015 x(10%4)m = 0.0004"
5.53 Im? of a sand-gravel aquifer, 201. of tetrachloroethylene, 20% of solubility,
a. From Table 5.12 the porosity is estimated at 0.20, from Table 5.15 the solubility is
given as 150 mg/L. Since the amount dissolved is only 20% ofits solubility:
20x 150 mg/l. x 1 m3 x 1034/3 x 0.20 = 6000 mg = 68
dissolved pere
b. Tofindreminingteachlroetylen, us th sp. gr. from Table 5.15 0F 1.63,
total mass=201 x 1.69 kyLx IB kg = 326003
undissolved mass = 32,600 632,594 g (almost ll not dissolved)
€: gradient = 0.001, porosity = 020,
average linear velocity = 2012 = SOME 3x 0.001 > 95 019. 1m/day
K(B4.) _ 410m/d x 0.001
7 020
a. using 1 m? cross section for oureubie meter of aquifer gives, andthe 6 gm?
Éontamhant concentration gives,
aquifer flow rate = 2.05 m/d x 1 m? = 2.05 mid
contaminant removal rate = 2.05 m3/d x 6 g/m? = 123 g/day
Lo remove the total 32,600 g would ta
pa 32,6008
123 /day x 3653ayiyr
=73 years
Pg 523
SOLUTIONS FOR CHAPTER 6
<1 Hardnessinmeg/L:
150mg/ L
20.05mg/ mea
29. SOmg/L
“12 15m8/meq
Total hardness = 7.5 + 4!
=7.5meq/L
Smeg tl.
124 meqiL.
Hardness as CaCO3:
Hardness #124684 y so BEMCICO, «mg/L a8 C200,
cs =
‘Table 6.5 would classify this as very hard water.
6.2
= 1x10 mol IL
x10%mol le 10'me mes soe,
HE, 1x10 meg IL
L “mol” g "Img
= 10° BE x50 200. 5410" mg/L as CACO,
+ 259 y sp ME CaCO, 0 5mg/ Las CaCO,
L men
ignoring the contribution of H* and OH, with pH=10.5, 39.0mg/L of
6.3 Fi
CO32- and 24.5 mg/L of HCO3”,
a cor ER) ¿op mE
E E
39.0mg/L , SOmBef CSCO, _ ¿omg LasCACO,
(co,
30m8/meg meq
Pg, 61
<1 Hardnessinmeg/L:
150mg/ L
20.05mg/ mea
29. SOmg/L
“12 15m8/meq
Total hardness = 7.5 + 4!
=7.5meq/L
Smeg tl.
124 meqiL.
Hardness as CaCO3:
Hardness #124684 y so BEMCICO, «mg/L a8 C200,
cs =
‘Table 6.5 would classify this as very hard water.
6.2
= 1x10 mol IL
x10%mol le 10'me mes soe,
HE, 1x10 meg IL
L “mol” g "Img
= 10° BE x50 200. 5410" mg/L as CACO,
+ 259 y sp ME CaCO, 0 5mg/ Las CaCO,
L men
ignoring the contribution of H* and OH, with pH=10.5, 39.0mg/L of
6.3 Fi
CO32- and 24.5 mg/L of HCO3”,
a cor ER) ¿op mE
E E
39.0mg/L , SOmBef CSCO, _ ¿omg LasCACO,
(co,
30m8/meg meq
Pg, 61
(4124309 - 61028
1 me
500;
=) = PASmelL_ Some of CaCO, „29 Img Las CaC
OO) gone me i
Ignoring H* and OH, alkalinity = (HCO3-) + (CO32)
220.1 +65.0=85.1 mg/L as CaCO3
b. Including H* and OH",
16x10" mal FL
ones, (H]=10""
(uy 2 ISuO ima, Le , Hh, Imes. , SOMECACO,
E ma "ms me
Sx10émg/Las CaCO,
16x10 mot /L.
3.16x10"mal _ 173, l0'mg me , SOmgCaCO,
(nr ióMO mal, 178, 1098 , q, , Soms CACO,
L mg “Umg ma
158 mg/L as CaCO,
alkalinity = (HCO3") + (CO32-) + (OH) -(H*)
11 + 65.0 + 158 - 1.510-6 = 100.9 mg/L as C2CO3
6.4 a.alkalinity = (HCO3") + (CO32*) + (OH!) - (H*),
gible,
but pH near neutral so (H*), (OH), and (CO3) are
16Smg/L_ SOmg of CaCO.
SE 352mg/Las CaCO,
omgImea mea,
(co; )=
alkalinity = 1352 mg/L as CaCOS
b. hardness = (Ca2*) + (Mg?)
of tata me, SOM, nn,
L “20mg” L *122m8)
€. total dissolved solids (TDS):
TDS =90 +30 + 72-464 120 + 225 + 165= 708 mg/L,
eh
Pg. 62
1 me
500;
=) = PASmelL_ Some of CaCO, „29 Img Las CaC
OO) gone me i
Ignoring H* and OH, alkalinity = (HCO3-) + (CO32)
220.1 +65.0=85.1 mg/L as CaCO3
b. Including H* and OH",
16x10" mal FL
ones, (H]=10""
(uy 2 ISuO ima, Le , Hh, Imes. , SOMECACO,
E ma "ms me
Sx10émg/Las CaCO,
16x10 mot /L.
3.16x10"mal _ 173, l0'mg me , SOmgCaCO,
(nr ióMO mal, 178, 1098 , q, , Soms CACO,
L mg “Umg ma
158 mg/L as CaCO,
alkalinity = (HCO3") + (CO32-) + (OH) -(H*)
11 + 65.0 + 158 - 1.510-6 = 100.9 mg/L as C2CO3
6.4 a.alkalinity = (HCO3") + (CO32*) + (OH!) - (H*),
gible,
but pH near neutral so (H*), (OH), and (CO3) are
16Smg/L_ SOmg of CaCO.
SE 352mg/Las CaCO,
omgImea mea,
(co; )=
alkalinity = 1352 mg/L as CaCOS
b. hardness = (Ca2*) + (Mg?)
of tata me, SOM, nn,
L “20mg” L *122m8)
€. total dissolved solids (TDS):
TDS =90 +30 + 72-464 120 + 225 + 165= 708 mg/L,
eh
Pg. 62
6.5 Questionable Cl”
loas mel menea mee
cat 90 200 45
M2+ 30 22 246
Ne 72 230 51
K 6 31 01
Toial= 1024
a (355 nass
So 225 480 469
HOO 165 610 270
Total = 739+ (CrY355
to balance, 102:
= 739 + (C355
CI-= 101 mg/l sothere probably was an error.
6.6 Wateranalysis
cat 95 200 475 215
Met 26 M2 213 1066
Net 15 20 065 326
HCOS" 160 éo 26 BL |
sog 135 480 281 140.6
ao ® 355 206 1028
a. total hardness (TH) = (Ca2+) + (Mg?*) = 237.5 + 1066 = 344.1 mg/L as CaCO3
b. carbonate hardness (CH) = (HCO3") = 131.1 mg/l. as CaCO3
=213 mgll as C2CO3
e. nonearbonate hardness (NCH) = 344.1 - 131
4, alkalinity =(HCO3"
e. TDS =95 +26 + 15 + 160+ 135 +73 = 504 mg/L
151.1 mg/L as CaCO3 (since pH near neutral)
Ps. 63
loas mel menea mee
cat 90 200 45
M2+ 30 22 246
Ne 72 230 51
K 6 31 01
Toial= 1024
a (355 nass
So 225 480 469
HOO 165 610 270
Total = 739+ (CrY355
to balance, 102:
= 739 + (C355
CI-= 101 mg/l sothere probably was an error.
6.6 Wateranalysis
cat 95 200 475 215
Met 26 M2 213 1066
Net 15 20 065 326
HCOS" 160 éo 26 BL |
sog 135 480 281 140.6
ao ® 355 206 1028
a. total hardness (TH) = (Ca2+) + (Mg?*) = 237.5 + 1066 = 344.1 mg/L as CaCO3
b. carbonate hardness (CH) = (HCO3") = 131.1 mg/l. as CaCO3
=213 mgll as C2CO3
e. nonearbonate hardness (NCH) = 344.1 - 131
4, alkalinity =(HCO3"
e. TDS =95 +26 + 15 + 160+ 135 +73 = 504 mg/L
151.1 mg/L as CaCO3 (since pH near neutral)
Ps. 63
a u A É$
ka a ——+
Th, aos
Æ werte
ws | = a
oe IT ss monaco
6.7 Wateranalysis:
1 mol. 2 yl
Gr 400 200 20 100.0
Mr 100 122 082 410
Nat X? 230 Y? Zn
Kt 70 391 0.18 9.0
ECOS 1100 610 18 902
sog 672 480 14 70.0
ar 10355 031 155
a. Equivalents balance to find Na concentration:
100 +41 + Z+9= 902 + 704 155 222537 mg/L as CaCOs
on, 257m8/ Las CaCO,
x2B3.0mg/meg = 1.8mplL
Somgimeg men
b. TH = (Ca2+) + (Mg2#) = 1000+ 41.0
141 mg/l. as CaCO3
nel:
m'y 3
ea
Ta
1257
k
CA SE [or
5 Er WEE 1757 mg/L es caos
6.8 Simplesaltbalance:
Sx106 Lid x 1500 mg/L
$x1500-3x75
2
15108 Lid x 75 mg/L + 2x106 Lidx C mg/L.
=3638mg/L
Pg, 64
ka a ——+
Th, aos
Æ werte
ws | = a
oe IT ss monaco
6.7 Wateranalysis:
1 mol. 2 yl
Gr 400 200 20 100.0
Mr 100 122 082 410
Nat X? 230 Y? Zn
Kt 70 391 0.18 9.0
ECOS 1100 610 18 902
sog 672 480 14 70.0
ar 10355 031 155
a. Equivalents balance to find Na concentration:
100 +41 + Z+9= 902 + 704 155 222537 mg/L as CaCOs
on, 257m8/ Las CaCO,
x2B3.0mg/meg = 1.8mplL
Somgimeg men
b. TH = (Ca2+) + (Mg2#) = 1000+ 41.0
141 mg/l. as CaCO3
nel:
m'y 3
ea
Ta
1257
k
CA SE [or
5 Er WEE 1757 mg/L es caos
6.8 Simplesaltbalance:
Sx106 Lid x 1500 mg/L
$x1500-3x75
2
15108 Lid x 75 mg/L + 2x106 Lidx C mg/L.
=3638mg/L
Pg, 64
6.9 Primary clarifier, overflow ate32m3/n?-day.
m
Mor rate _ 200009 _ gan? q
Overtow rate = HOW Rte - 2000m de
day
2000
Le 2000 156m
432
volume _ 2.4m x dm x 156m x 24hr/day
Detention ime = AE. 20000 day
=18 hr
6.10 2mgd, SOO0g/4-2, DT>2 hr, deptho 1 fl, circular,
SE ners
ES
o
ee |
own. 20 pi/é
Seelow rie” BOOgal d=
Detention time constraint
pr = Llene 2508 <I _ MR ate
iow Fate 2519 gal dx. y
2x10"gal /d Tagg
Dz: 2 8
aa
Minimum depth constrain says depth must be a least 11 ft(more than the detention
time constraint). Therefore depth D = 11 ft
Pg. 65
m
Mor rate _ 200009 _ gan? q
Overtow rate = HOW Rte - 2000m de
day
2000
Le 2000 156m
432
volume _ 2.4m x dm x 156m x 24hr/day
Detention ime = AE. 20000 day
=18 hr
6.10 2mgd, SOO0g/4-2, DT>2 hr, deptho 1 fl, circular,
SE ners
ES
o
ee |
own. 20 pi/é
Seelow rie” BOOgal d=
Detention time constraint
pr = Llene 2508 <I _ MR ate
iow Fate 2519 gal dx. y
2x10"gal /d Tagg
Dz: 2 8
aa
Minimum depth constrain says depth must be a least 11 ft(more than the detention
time constraint). Therefore depth D = 11 ft
Pg. 65
SOLUTIONS FOR CHAPTER 7
7. From(18), (at | aim and 25°C)
5000ppm x (12+ 2x16)
a CO,mg/ m? = =8%92mg/m = 9000mg/ m°
CO,mg/ a ang "
24.465 x36 man? ou
a tm
LO m/m à 2500 (14416) >
e. NO mg/m! = PR CLIS) 2 50.7mg/m
NO mg BAUS) 307 me
7.2. 10% efficient scrubber, find S emission rate:
600/0.38=1879 mt soo me
SO venons eae
9000 Bru coal 19 8
600.000 kWe 3412Biw_ Ibcoal 00LIbS
038 kWhr "90008
= 5986 Ib Sihr
70% efficient, says release 0.3 x 5986 IbS/hr = 1796 Ib S/hr =1800 IbS/hr
7.3 [fall converted to $02 and now using a 90% efficient scrubber:
5986 IS (32-+2x16) 1b SO.
$0,=0.1x es
19716 SO, / hr = 12001b SO, /hr
7.4 70% scrubber, 0.6 Ib SO2/106 Btu in, find % S allowable:
15,000x06
CETTE
5% S fuel
X Ibs , 03 los Sout 210950, , Ibesal
ibscoal * 11bSin ~ IbS "3,0008
0.009 0.9% S fuel
Pg. 7.1
7. From(18), (at | aim and 25°C)
5000ppm x (12+ 2x16)
a CO,mg/ m? = =8%92mg/m = 9000mg/ m°
CO,mg/ a ang "
24.465 x36 man? ou
a tm
LO m/m à 2500 (14416) >
e. NO mg/m! = PR CLIS) 2 50.7mg/m
NO mg BAUS) 307 me
7.2. 10% efficient scrubber, find S emission rate:
600/0.38=1879 mt soo me
SO venons eae
9000 Bru coal 19 8
600.000 kWe 3412Biw_ Ibcoal 00LIbS
038 kWhr "90008
= 5986 Ib Sihr
70% efficient, says release 0.3 x 5986 IbS/hr = 1796 Ib S/hr =1800 IbS/hr
7.3 [fall converted to $02 and now using a 90% efficient scrubber:
5986 IS (32-+2x16) 1b SO.
$0,=0.1x es
19716 SO, / hr = 12001b SO, /hr
7.4 70% scrubber, 0.6 Ib SO2/106 Btu in, find % S allowable:
15,000x06
CETTE
5% S fuel
X Ibs , 03 los Sout 210950, , Ibesal
ibscoal * 11bSin ~ IbS "3,0008
0.009 0.9% S fuel
Pg. 7.1
7.9
zu
RO + +0, > HO, »+R'CHO 19)
for R'CHO to be HCHO, R' must be H so that
RO «+0, >HO, + +HCHO
forthereactiontobalance . R = CH,
Which says RH in (7.16) must be CH, (methane)
RH = propene = CH2=CH-CH3 =C3H6 so, R=C3H5
so the sequence of reactions (7.16) 0 (7.19) are:
CH, +OHe > GH, #10
GH, +0, 9 CH0,+
H,O, ++NO >C,H.0 ++NO,
CH,0 + +0, + HO, »+C¿H,CHO
‘The end product is acrolein, CH2CHCHO.
LAS overt
tema = essen 2002 10000
Bw
BY - 13700'"Bm
[2
kWh
0349 =35%
ALNSPS of 0.03 Ib particulates per 106 Btu input, emissions would have been:
0.08 Ib
in x 10008
Sp x 13710 Bu in x 87x10
TO Bru heat input ETS
emissions at NSPS _ 1.87x10'g
comparison, SmuSSIONS aL NSPS | LE = 0.48 = 48%
Fon cal emo 0310 =O
Derivation forthe dry adiabatic lapse rate
4Q=dU+dW where dU =
dQ=C,A+PIV (m
ideal gas law says PV = nT
and dW = Pav
so, d(PV) = PaV + VaP = 2RT
or, PAV=nRT-V4P
plugged into(1) gives:
Ps. 73
zu
RO + +0, > HO, »+R'CHO 19)
for R'CHO to be HCHO, R' must be H so that
RO «+0, >HO, + +HCHO
forthereactiontobalance . R = CH,
Which says RH in (7.16) must be CH, (methane)
RH = propene = CH2=CH-CH3 =C3H6 so, R=C3H5
so the sequence of reactions (7.16) 0 (7.19) are:
CH, +OHe > GH, #10
GH, +0, 9 CH0,+
H,O, ++NO >C,H.0 ++NO,
CH,0 + +0, + HO, »+C¿H,CHO
‘The end product is acrolein, CH2CHCHO.
LAS overt
tema = essen 2002 10000
Bw
BY - 13700'"Bm
[2
kWh
0349 =35%
ALNSPS of 0.03 Ib particulates per 106 Btu input, emissions would have been:
0.08 Ib
in x 10008
Sp x 13710 Bu in x 87x10
TO Bru heat input ETS
emissions at NSPS _ 1.87x10'g
comparison, SmuSSIONS aL NSPS | LE = 0.48 = 48%
Fon cal emo 0310 =O
Derivation forthe dry adiabatic lapse rate
4Q=dU+dW where dU =
dQ=C,A+PIV (m
ideal gas law says PV = nT
and dW = Pav
so, d(PV) = PaV + VaP = 2RT
or, PAV=nRT-V4P
plugged into(1) gives:
Ps. 73
Q= CAT + nRAT - VaP
cmo vie
RV O
at constant pressure :
Bec rc,
a
putting that into (2) gives,
eQ=CjaT-vaP whict
7.13 Plotting the data, extending from groundlevel to crossing with amt
lapse rate, and extending from the stack height gives:
profile at the adiabatic
800
a m i =
2 king death | > Bal
$ 0
3 «lt
nen
o 111 | =
vane EN E
Temperature (€)
2) mixing depth (projecting from 200C at Om at slope -10/100m) = 400 m
») plume rise (projecting from 219C at 100m) = 500m
7.14 Projection from the ground at 22°C crosses ambient at 500m.
Need the windspeed at 250 m (halfway up) using (7.43) and Table 7.7 for Class C,
4 (BY he (m
E © m/s” om
Us =1.90K4 = 7.6/5
Ventilation coeff = 500m x 7.6m/ = 3.8x103 m2s
Pa 74
cmo vie
RV O
at constant pressure :
Bec rc,
a
putting that into (2) gives,
eQ=CjaT-vaP whict
7.13 Plotting the data, extending from groundlevel to crossing with amt
lapse rate, and extending from the stack height gives:
profile at the adiabatic
800
a m i =
2 king death | > Bal
$ 0
3 «lt
nen
o 111 | =
vane EN E
Temperature (€)
2) mixing depth (projecting from 200C at Om at slope -10/100m) = 400 m
») plume rise (projecting from 219C at 100m) = 500m
7.14 Projection from the ground at 22°C crosses ambient at 500m.
Need the windspeed at 250 m (halfway up) using (7.43) and Table 7.7 for Class C,
4 (BY he (m
E © m/s” om
Us =1.90K4 = 7.6/5
Ventilation coeff = 500m x 7.6m/ = 3.8x103 m2s
Pa 74
7.18 Below the knee, the plume is fanning which suggests a stable atmosphere, which could be
profile (a), () or td).
‘Above the knee, the plume is looping, which suggests superadiabatic, which isd
7.16 Projecting the dry adiabatic lapse rate, and the dew point lapse rate, gives:
4000 T
[ [saturated sabbalie|
uam
3000 :
¿ end
< dl
jo tot
& N
EE SER:
RECENSE
TIIITINII
oki
OUTRE SOIT 14 1610 20 20 242520 30.32 36
Temperature (C)
a. crossover point is where clouds begin o form (saturated ai) = 2000
bo at km, the temperature would be 100 - (69/km x Ikm) =4 °C.
€: falling from 3km at 4°C and incrasing 100C per km, reaches 4+3x10=349C
7.17 A 201m article blown to 8000 m, setling velocity,
For _ 20xi0%m)*x 1.5410'g/ x 9.800
187 18 x O01 72g/m-s
aus Be
0,019m 15% 3600./hr x Zährid
Time to reach the ground =
horizontal distance = 4.87 days x 10 m/s x 3600s/hr x 24hr/d x 10°3km/m = 4200 km
7.18 Residence time for 10 um particle, unit density, at 1000m:
„Cor _ 1x10" m)*x 10%/m' x9:80m/F
= =0.00317m/s |
18 18 K00172g/m-s
settling velocity
Pg. 75
profile (a), () or td).
‘Above the knee, the plume is looping, which suggests superadiabatic, which isd
7.16 Projecting the dry adiabatic lapse rate, and the dew point lapse rate, gives:
4000 T
[ [saturated sabbalie|
uam
3000 :
¿ end
< dl
jo tot
& N
EE SER:
RECENSE
TIIITINII
oki
OUTRE SOIT 14 1610 20 20 242520 30.32 36
Temperature (C)
a. crossover point is where clouds begin o form (saturated ai) = 2000
bo at km, the temperature would be 100 - (69/km x Ikm) =4 °C.
€: falling from 3km at 4°C and incrasing 100C per km, reaches 4+3x10=349C
7.17 A 201m article blown to 8000 m, setling velocity,
For _ 20xi0%m)*x 1.5410'g/ x 9.800
187 18 x O01 72g/m-s
aus Be
0,019m 15% 3600./hr x Zährid
Time to reach the ground =
horizontal distance = 4.87 days x 10 m/s x 3600s/hr x 24hr/d x 10°3km/m = 4200 km
7.18 Residence time for 10 um particle, unit density, at 1000m:
„Cor _ 1x10" m)*x 10%/m' x9:80m/F
= =0.00317m/s |
18 18 K00172g/m-s
settling velocity
Pg. 75
7.21
1000m
IM ___- 57.6
A 3600s 8
residence time
Settling velocity and Reynolds numbers:
pp _ ARO mx 10'g/m x 980M Là 2 197,
a tum: v= T2 18x 001729m-+
re = usd Ola lO m3 170 ma =24x10%
SNS 2400
ONO mpx 10%g/ a? 19.800 _ o 917715
18x 0.0172g)m-s
1.29x10° g/m? x 20x10%m x 0.0127m/s
0.0172 ms
02
H=SOm, overcast so Class D, A at 1.2km, B at 1 km:
a. Fig 7.50, Class D, H=50m, max concentration at Ikm. Since concentration is decreasing
‘past À km, the higher level of pollution will beat site "A".
b. Clear sky, wind < Sm/s: Classis now A,B or C. At Som, Class A.B, or C, Fig 7.50
Shows us that the maximum point moves closer to the stack.
+. Itwill stil be house at site “A.”
Bonfire, 20g/s CO, wind 2 m/s, H=6m, distanc
‘Table 7.8, clear night, stability elassificatio
x. Le
a. At400m, oy
20x10 ug/s
minima HT)
= 2x0 pg/m? = 2img/ n°
b. Atthe maximum point, Fig. 7.50
Ps. 76
1000m
IM ___- 57.6
A 3600s 8
residence time
Settling velocity and Reynolds numbers:
pp _ ARO mx 10'g/m x 980M Là 2 197,
a tum: v= T2 18x 001729m-+
re = usd Ola lO m3 170 ma =24x10%
SNS 2400
ONO mpx 10%g/ a? 19.800 _ o 917715
18x 0.0172g)m-s
1.29x10° g/m? x 20x10%m x 0.0127m/s
0.0172 ms
02
H=SOm, overcast so Class D, A at 1.2km, B at 1 km:
a. Fig 7.50, Class D, H=50m, max concentration at Ikm. Since concentration is decreasing
‘past À km, the higher level of pollution will beat site "A".
b. Clear sky, wind < Sm/s: Classis now A,B or C. At Som, Class A.B, or C, Fig 7.50
Shows us that the maximum point moves closer to the stack.
+. Itwill stil be house at site “A.”
Bonfire, 20g/s CO, wind 2 m/s, H=6m, distanc
‘Table 7.8, clear night, stability elassificatio
x. Le
a. At400m, oy
20x10 ug/s
minima HT)
= 2x0 pg/m? = 2img/ n°
b. Atthe maximum point, Fig. 7.50
Ps. 76
7.23
7.24
(&) 38x10’ Im’
_ 20x10? mgs, 3.8x10%
=38 = 40mg! m"
Zmis m Be
1=100m, Q=1.2g/s, wH=4m/s, us=3*mís, C<365ugm3.
“The more unstable the atmosphere, the higher the peak downwind concentration (see Fig,
7.49). From Table 7.8, with wind > 3m/s,B is the most unstable so it leads to the worst
concentration:
from Hg. 750, Xp 0m (Sa) 1500970
Su0*
cu,
Stu} 236510"
Q L il
„4365810“
15107
AE
Mw
Maximum power plant size = 97 gis x => = 8OMW
128%
Atmospheric conditions, stack height, and groundlevel concentration restrictions same as
Prob. 7.22 so hat
Emissions Q = 97 y/s
_OSIDSO, 1Btwin „3412Biuout, Ihr 10%
52 awe Aue,
El TETU 2210" 0"
9IXIO'x035:3600x2:2
Por
: 0.3412x1000,
31.000KW =130MW
H=100m, ua=4 m/s, Q=80g/s, clear summer day so Class B:
First, find the windspeed at the effective sack height using (7.43) and Table 7.7:
DE
y
7.24
(&) 38x10’ Im’
_ 20x10? mgs, 3.8x10%
=38 = 40mg! m"
Zmis m Be
1=100m, Q=1.2g/s, wH=4m/s, us=3*mís, C<365ugm3.
“The more unstable the atmosphere, the higher the peak downwind concentration (see Fig,
7.49). From Table 7.8, with wind > 3m/s,B is the most unstable so it leads to the worst
concentration:
from Hg. 750, Xp 0m (Sa) 1500970
Su0*
cu,
Stu} 236510"
Q L il
„4365810“
15107
AE
Mw
Maximum power plant size = 97 gis x => = 8OMW
128%
Atmospheric conditions, stack height, and groundlevel concentration restrictions same as
Prob. 7.22 so hat
Emissions Q = 97 y/s
_OSIDSO, 1Btwin „3412Biuout, Ihr 10%
52 awe Aue,
El TETU 2210" 0"
9IXIO'x035:3600x2:2
Por
: 0.3412x1000,
31.000KW =130MW
H=100m, ua=4 m/s, Q=80g/s, clear summer day so Class B:
First, find the windspeed at the effective sack height using (7.43) and Table 7.7:
DE
y
CE es
moja, 20,
ca sono 23
= 5.65m/s x290m x 234m
(x,0)=;
b. Atthe maximum point, 0.7 km (Fig. 7.50),
(Se) = nssot rat
26.
km, y=0.1km:
e At
A eu
70,9,
fa
coreo E
om jogo
ART
So, assume oy =k Gz, then
a. 0) en
a. 0 |
%, rok [oe
multiply through by 625 and cancel lots of terms to get,
Ei
20; or 6,27 070
b. substituting the newly found value for 02,
Ps. 78
moja, 20,
ca sono 23
= 5.65m/s x290m x 234m
(x,0)=;
b. Atthe maximum point, 0.7 km (Fig. 7.50),
(Se) = nssot rat
26.
km, y=0.1km:
e At
A eu
70,9,
fa
coreo E
om jogo
ART
So, assume oy =k Gz, then
a. 0) en
a. 0 |
%, rok [oe
multiply through by 625 and cancel lots of terms to get,
Ei
20; or 6,27 070
b. substituting the newly found value for 02,
Ps. 78
2
Fos,
e. using Oy =k oz
42)
0.117
‘oR(O-TO7HY
7.26 Sudbury stack:
130°C À zoms — 10%
15.20% sem
{Using (7.49) for bouyancy flux parameter
Fx 10 +273 “Te
x20 {1 10273.) 3370/9
Joo pes)
and distance downwind to final plume rise xf given on page 420 (with F>55),
Xp = 120F** = 120x(3370)"* =3092
for stability classification C, use (751) for plume rise,
LEFUx _ 1.6x(3370)!(3092)"
> 8
=635m plume ise
H= effective stack height=h + Ah = 380 + 635 = 1015m
7.27 Repeat P7.26 with a stable, isothermal atmosphere:
F = 3370 múis3 from Prob. 7.26, but for isothermal atmosphere, need the stability
(8 sone) = 288 CE
which is used in (7.48) for plume rise under these conditions,
nea E" ed ate)”
z) =278m
Sais x 3.461078
Ps. 79
Fos,
e. using Oy =k oz
42)
0.117
‘oR(O-TO7HY
7.26 Sudbury stack:
130°C À zoms — 10%
15.20% sem
{Using (7.49) for bouyancy flux parameter
Fx 10 +273 “Te
x20 {1 10273.) 3370/9
Joo pes)
and distance downwind to final plume rise xf given on page 420 (with F>55),
Xp = 120F** = 120x(3370)"* =3092
for stability classification C, use (751) for plume rise,
LEFUx _ 1.6x(3370)!(3092)"
> 8
=635m plume ise
H= effective stack height=h + Ah = 380 + 635 = 1015m
7.27 Repeat P7.26 with a stable, isothermal atmosphere:
F = 3370 múis3 from Prob. 7.26, but for isothermal atmosphere, need the stability
(8 sone) = 288 CE
which is used in (7.48) for plume rise under these conditions,
nea E" ed ate)”
z) =278m
Sais x 3.461078
Ps. 79
H
fective stack height = h + Ah = 380 + 278
(notice the atmospheric stability lowered effective stack height vs Prob. 7.26)
7.28 Cloudy summer day, stability classification C (Table 78),
moho ms —» 60%
Æ Sms
ma 1
100m
Using (7.49) for bouyancy flux parameter
da 64273
e, Jante
1204273
and distance downwind to final plume rise xr given on page 420 (with F<55),
x, = SOP = 50x(28.4)" =406
for stability classification C, use (7.51) for plume rise,
REE
= S4m plume rise
frective stack height = h-+ Ah = 100+ $4= 154m
7.29 Power plant, find groundlevel pollution 16 km away. Need fist find H.
Q300p 502 TE Cass E
— sms
sas lapse rate» Ci
18500,
100m
ree sm
200 MWe on,
First, find bouyaney flux parameter (7.49),
ÿ as
145+27
plume rise for stable (Class E) atmosphere needs $ from (7.50),
Pg. 7.10
fective stack height = h + Ah = 380 + 278
(notice the atmospheric stability lowered effective stack height vs Prob. 7.26)
7.28 Cloudy summer day, stability classification C (Table 78),
moho ms —» 60%
Æ Sms
ma 1
100m
Using (7.49) for bouyancy flux parameter
da 64273
e, Jante
1204273
and distance downwind to final plume rise xr given on page 420 (with F<55),
x, = SOP = 50x(28.4)" =406
for stability classification C, use (7.51) for plume rise,
REE
= S4m plume rise
frective stack height = h-+ Ah = 100+ $4= 154m
7.29 Power plant, find groundlevel pollution 16 km away. Need fist find H.
Q300p 502 TE Cass E
— sms
sas lapse rate» Ci
18500,
100m
ree sm
200 MWe on,
First, find bouyaney flux parameter (7.49),
ÿ as
145+27
plume rise for stable (Class E) atmosphere needs $ from (7.50),
Pg. 7.10
ES
sop (vo) (con)
plume rise is given by (7.48),
e w
F ESTO
so the effective heightis H= 100m + [21m =221 m |
Concentration downwind at 16km: (Table 7.10) oy = 602m, 02 = 95m
ae
sooo s
ae
FE om (205)
7.30
+ mas
sms NT
her [T
(Cu, 20x10*ug/s{ Cu, Cu, Se
Conn (Se) 2ox1o%ug/s( Con | uE Is +
uw Q Smis Q Q
Using Fig. 7.50 gives,
Cas) sono 0.25km, Coy, = 4x10" x 610
(
Ve
(
\
240 pg! m’
A
seco 0.55km. Coy’ 4x10° x 58x10*=23048/00”
96ug/m"
Q
=) 24:10 % 3.7, Ca. = 410" x 24X10"
\e
Peru
sop (vo) (con)
plume rise is given by (7.48),
e w
F ESTO
so the effective heightis H= 100m + [21m =221 m |
Concentration downwind at 16km: (Table 7.10) oy = 602m, 02 = 95m
ae
sooo s
ae
FE om (205)
7.30
+ mas
sms NT
her [T
(Cu, 20x10*ug/s{ Cu, Cu, Se
Conn (Se) 2ox1o%ug/s( Con | uE Is +
uw Q Smis Q Q
Using Fig. 7.50 gives,
Cas) sono 0.25km, Coy, = 4x10" x 610
(
Ve
(
\
240 pg! m’
A
seco 0.55km. Coy’ 4x10° x 58x10*=23048/00”
96ug/m"
Q
=) 24:10 % 3.7, Ca. = 410" x 24X10"
\e
Peru
240
28
amd)
ss
025 085
Ts
LEIOA an un a ve
BROS EIN
ker
ine He 2,
@ Son: Cmax= 41106 x 57:10 =228 ugm?
@ 100m: Cmax = 4106 x LSKIOS = 6Op gin?
@200m: Cmax = 44108 x 3.4x106 = 14 wand
2 an) 1
Do they drop as (WH)? that is expectation : LED
cm 4
200m) 1
testthem SUR 6
sm) ” 28
(200m) _ 1 (200m) _ 14
expect SEP J um). 0061
ae C(S0m) 16 un: C(S0m) 228
7.32. Papermill emitting H2S, Ikm away want 0.1 x odor threshold:
ña
Pg. 712
A m
am" ge 7028 not bad!
not bad.
28
amd)
ss
025 085
Ts
LEIOA an un a ve
BROS EIN
ker
ine He 2,
@ Son: Cmax= 41106 x 57:10 =228 ugm?
@ 100m: Cmax = 4106 x LSKIOS = 6Op gin?
@200m: Cmax = 44108 x 3.4x106 = 14 wand
2 an) 1
Do they drop as (WH)? that is expectation : LED
cm 4
200m) 1
testthem SUR 6
sm) ” 28
(200m) _ 1 (200m) _ 14
expect SEP J um). 0061
ae C(S0m) 16 un: C(S0m) 228
7.32. Papermill emitting H2S, Ikm away want 0.1 x odor threshold:
ña
Pg. 712
A m
am" ge 7028 not bad!
not bad.
7.33
age EN
)
110" g/m? — —
DO gm = nm x Om ZO”
Mao à =
ru 156x 110x001K0
ox [2420010 742)"
so, at each end ofthe wind speed range we can find the height needed:
u. [raon(22] ac
Hane 2420010(242)]" «20m
If the town extends beyond 1 km, from Fig 7.50 et H=265, Class B, Xmax =1.8km
says to be conservative use H=265m
‘Therefore, with the peak occurring beyond the I km house, the concentration will ise
for buildings located > Ikm away. YES
Stack under an inversion:
150 gs
XL
atx= XL 622 047 (L-H) = 047 (100 - 45) = 26 m
(orcas. op 26m at x= 0m (able 7.10), heros XL = 04 km, and lo
from Table 7.10, oy =
L: C(K,,0)=
axes mu | 2%,
150x10’mg/s
FS ns x 46m x 26m
atx=2XL: oy = 85m (Table 7.10)
Pg. 7.13
age EN
)
110" g/m? — —
DO gm = nm x Om ZO”
Mao à =
ru 156x 110x001K0
ox [2420010 742)"
so, at each end ofthe wind speed range we can find the height needed:
u. [raon(22] ac
Hane 2420010(242)]" «20m
If the town extends beyond 1 km, from Fig 7.50 et H=265, Class B, Xmax =1.8km
says to be conservative use H=265m
‘Therefore, with the peak occurring beyond the I km house, the concentration will ise
for buildings located > Ikm away. YES
Stack under an inversion:
150 gs
XL
atx= XL 622 047 (L-H) = 047 (100 - 45) = 26 m
(orcas. op 26m at x= 0m (able 7.10), heros XL = 04 km, and lo
from Table 7.10, oy =
L: C(K,,0)=
axes mu | 2%,
150x10’mg/s
FS ns x 46m x 26m
atx=2XL: oy = 85m (Table 7.10)
Pg. 7.13
C(2X,,0)
Q 150x10°mg/s A
le oe
Tr Te "48
7.34 Stack underaninversion ayer:
AAN +
re Le2s0n
som || iy
ellas: de
XL pr
We need the stability classification: clear summer day, dm/s, Table 7.8 says Class B.
atx=XL 02=047(L-H)=047(250 - 50) = 94m
a. From Table 7.10, at 07 = 94m Class B, XL =09km. Since our point of interest is
“at 4 km, we are well past the point at which reflections first occur so we can use
(7.52). We need oy at km, which is given in Table 7.10 as 539m:
Q BOxI0‘ug/s >
(ákm.0)= IL
eu) Fr vo,L ar x Smisx 539m x 250m Ar
. Without the inversion layer, at km 07 = 498m, Gy = 539m so,
[.
me
7.35 Agricultural bum
Clakm,0)
198 1m
soxiotugis EA
FR Sms x 539m x 458m | DRA
0.30/50 p
D
Sams
7
Clear fall afternoon, winds 3 m/s, so stability class °C” (Table 7.8),
and 67 = 26m (Table 7.10). Using (7.54),
brn) I 20m MS 078100
Cloak) erg,” Jan x amin 2am Oe
Pg. 7.14
Q 150x10°mg/s A
le oe
Tr Te "48
7.34 Stack underaninversion ayer:
AAN +
re Le2s0n
som || iy
ellas: de
XL pr
We need the stability classification: clear summer day, dm/s, Table 7.8 says Class B.
atx=XL 02=047(L-H)=047(250 - 50) = 94m
a. From Table 7.10, at 07 = 94m Class B, XL =09km. Since our point of interest is
“at 4 km, we are well past the point at which reflections first occur so we can use
(7.52). We need oy at km, which is given in Table 7.10 as 539m:
Q BOxI0‘ug/s >
(ákm.0)= IL
eu) Fr vo,L ar x Smisx 539m x 250m Ar
. Without the inversion layer, at km 07 = 498m, Gy = 539m so,
[.
me
7.35 Agricultural bum
Clakm,0)
198 1m
soxiotugis EA
FR Sms x 539m x 458m | DRA
0.30/50 p
D
Sams
7
Clear fall afternoon, winds 3 m/s, so stability class °C” (Table 7.8),
and 67 = 26m (Table 7.10). Using (7.54),
brn) I 20m MS 078100
Cloak) erg,” Jan x amin 2am Oe
Pg. 7.14
7.36 A freeway modelled asa line source:
water 4 |
1,5 g/mi
Clear summer, 2 ms, Table 7.8 suggests Class A or A-B,
Im for Class B. What should we use? Since
> itis somewhere between Class A and Class B, but closer to A, les use 07 = 26m:
Tofind the linear emission rate:
RC NU RS mi
1! 36005” mi — vehicle "52801 * 0.3048m
Casos ms =2.8mg/m=s
Then, using (7.54),
2a _ 2258me/m
=0.04mg/m’
Toro, Tor x2misx 2m À
C(2km)
7.37 Box model, 250,000 vehicles between 4 and 6pm, driving 40km ea, emitting 4g/km CO.
km 4800 1, he 1 secon
PEUR = 46x10%gC0/ m's
veh km 'Zhrs 36005 15x80x10°m" aan
b. Using (7.58) with t= 2hrs x 3600s/he =
a. 9, =250,000veh. x:
200s,
EN)
‘6nl0g/ms x 15000m
sn x 150000, )= 00028/m? =21g/
+. With no wind, go back to (7.55) and solve the differential equation:
ES
pr = UV
ac
D
Cain Ax10*gCO/m
15m
= 00022300 /m’ = 2.2mg/m*
Pa. 715
water 4 |
1,5 g/mi
Clear summer, 2 ms, Table 7.8 suggests Class A or A-B,
Im for Class B. What should we use? Since
> itis somewhere between Class A and Class B, but closer to A, les use 07 = 26m:
Tofind the linear emission rate:
RC NU RS mi
1! 36005” mi — vehicle "52801 * 0.3048m
Casos ms =2.8mg/m=s
Then, using (7.54),
2a _ 2258me/m
=0.04mg/m’
Toro, Tor x2misx 2m À
C(2km)
7.37 Box model, 250,000 vehicles between 4 and 6pm, driving 40km ea, emitting 4g/km CO.
km 4800 1, he 1 secon
PEUR = 46x10%gC0/ m's
veh km 'Zhrs 36005 15x80x10°m" aan
b. Using (7.58) with t= 2hrs x 3600s/he =
a. 9, =250,000veh. x:
200s,
EN)
‘6nl0g/ms x 15000m
sn x 150000, )= 00028/m? =21g/
+. With no wind, go back to (7.55) and solve the differential equation:
ES
pr = UV
ac
D
Cain Ax10*gCO/m
15m
= 00022300 /m’ = 2.2mg/m*
Pa. 715
7.38 Box model, 105 m on a side, H=1200m, u=4m/s, SO2=20kg/s, steady state:
Amis,
Cine a
20k 9/5
ine enge
O A
20x10”
zos
4178/00
7.39 Assume steady-state conditions were achieved by Spm Friday o that C(0)=41.7h.g/m°
With qs = 0, end Cin
(7ST gives us Cd = COJE"
a. At midnight,
= Clo)" = A1 Tag me ce" ENE à
saut
b. Starting up again at Sam on Monday, by Spm (Shrs tater):
{first check to see concentration left from Friday at Spm (63 hrs earlier):
C(t) = ClO}e™™"* = 41.7ygh m? «etm ODES
00Syg/m’ =0
so we can ignore that and Jet C(O) = Oat Sam Monday. First find the emissions per
unitarea,
emission rate _ 20kg/sx 10° ng/kg a:
en D
“Then use (7.57) with Cin =0%
oe
oi )
206) mx 108m, aie me soon) 2
sim 1 = 3024 /
Pg. 7.16
Amis,
Cine a
20k 9/5
ine enge
O A
20x10”
zos
4178/00
7.39 Assume steady-state conditions were achieved by Spm Friday o that C(0)=41.7h.g/m°
With qs = 0, end Cin
(7ST gives us Cd = COJE"
a. At midnight,
= Clo)" = A1 Tag me ce" ENE à
saut
b. Starting up again at Sam on Monday, by Spm (Shrs tater):
{first check to see concentration left from Friday at Spm (63 hrs earlier):
C(t) = ClO}e™™"* = 41.7ygh m? «etm ODES
00Syg/m’ =0
so we can ignore that and Jet C(O) = Oat Sam Monday. First find the emissions per
unitarea,
emission rate _ 20kg/sx 10° ng/kg a:
en D
“Then use (7.57) with Cin =0%
oe
oi )
206) mx 108m, aie me soon) 2
sim 1 = 3024 /
Pg. 7.16
7-40 Steady-state conditions from Prob 7.38, wind drops to 2 ms, 2hes later:
From Prob. 7.38, emission rate qs = 2.0 jig/m?-s, and C = 41.7ug/m3, Using (7.57),
ct) si =e Coe
om ne ae hd
POE TITS
some)
Amex 12000 )
J+ 41.76"
=473 um?
i 7.41. Modified Prob. 738,now incoming aras Sim? and there are 10 gn? already
thereat am, Find the concertos at none
ae.
os
£
1200m
om T
Tom
20 gms
} at
Clann) ETES sara crea)
“4m/sx 1200m
gen ena)
‘C(4hr = noon) = 26.1 4g/m3.
7.42. Now wing ondes of rob 738 tl for nenne pollu with 2023
Ratna ox Rate toto Rate of deay
S=uWHe + Kev
CI (us
PUR um om 0omc{"¥/3)
«(92 cH not
hrm? “36005,
10% x 10'mx 1200m }
20x10? = 481108 € + 76.108
© = 164gm3
Pg. 717
From Prob. 7.38, emission rate qs = 2.0 jig/m?-s, and C = 41.7ug/m3, Using (7.57),
ct) si =e Coe
om ne ae hd
POE TITS
some)
Amex 12000 )
J+ 41.76"
=473 um?
i 7.41. Modified Prob. 738,now incoming aras Sim? and there are 10 gn? already
thereat am, Find the concertos at none
ae.
os
£
1200m
om T
Tom
20 gms
} at
Clann) ETES sara crea)
“4m/sx 1200m
gen ena)
‘C(4hr = noon) = 26.1 4g/m3.
7.42. Now wing ondes of rob 738 tl for nenne pollu with 2023
Ratna ox Rate toto Rate of deay
S=uWHe + Kev
CI (us
PUR um om 0omc{"¥/3)
«(92 cH not
hrm? “36005,
10% x 10'mx 1200m }
20x10? = 481108 € + 76.108
© = 164gm3
Pg. 717
D oQ o eee
7.43 Starting with (761) and using the special conditions ofthis tracer-gas study; that is, a
‘conservative tracer (K=0),no racer in the air leaking into the room (Ca=0), and the tracer
source tured off at 1=0 ($=0) gives the exponential decay of tracer as
simeghe) _C (ppm) IC
o 100 2303
0.5 80 209
12 60 1m
15 50 16
20 33 1194
me
time (he)
From the graph, the slope is about: slope =
‘Thus, the infiltration rate is about 0.53 air changes per hour.
7.44 Infiltration 0 Sach, 500m? volume, 200 m? floor space, radon 0.6pCi/m2:
Using (7.60) with K=7.6x10-3/hr (Table 7.15),
ey Los
T+K (0.5/hr+7.6x107 /hr}x
}700PCi Fen? =1:TPCi/L
Ps, 718
7.43 Starting with (761) and using the special conditions ofthis tracer-gas study; that is, a
‘conservative tracer (K=0),no racer in the air leaking into the room (Ca=0), and the tracer
source tured off at 1=0 ($=0) gives the exponential decay of tracer as
simeghe) _C (ppm) IC
o 100 2303
0.5 80 209
12 60 1m
15 50 16
20 33 1194
me
time (he)
From the graph, the slope is about: slope =
‘Thus, the infiltration rate is about 0.53 air changes per hour.
7.44 Infiltration 0 Sach, 500m? volume, 200 m? floor space, radon 0.6pCi/m2:
Using (7.60) with K=7.6x10-3/hr (Table 7.15),
ey Los
T+K (0.5/hr+7.6x107 /hr}x
}700PCi Fen? =1:TPCi/L
Ps, 718
0.6pc/m2 s
Ka7.6x10°3 e
7.45 Just have half as much ground-floor area to let radon i
1.6 pCi/m*s x 100r
(05/8 +7.6300° /he)x E
7m? = 085pCi/ L
7.46 Using exposure factors from Table 4.10 and potency from Prob. 4.10 and 4.12,
17pCi1L x360%%.x30yr
SEES yee ae
ose = = 0-72pCi Laverage over 70 yrs
Esp Erz HQE se er 70.
oy „ One yt
PORT SOC
Risk =0:72pCi/ Lx Qomem/ ye, Leaneer death , rem cy 200017 = 0.2%
TSPCi/L “7 8000rem TO’ meer
7.47 20013 house, 0 2ach, oven+2bumers 6pm to Tpm, find CO pm and and 10pm, For
‘these circumstances, (7.62) is appropriate:
Sy ge
CW = Fle")
the source strength, Sis
67pm: oven + 2 bumers,1900mphr= 251540m/r = 5580mghr
solving for Cafter he:
(Up) = SOE NN) 16.8mg/
02 HEHEHE om”
Pe. 7.19
Ka7.6x10°3 e
7.45 Just have half as much ground-floor area to let radon i
1.6 pCi/m*s x 100r
(05/8 +7.6300° /he)x E
7m? = 085pCi/ L
7.46 Using exposure factors from Table 4.10 and potency from Prob. 4.10 and 4.12,
17pCi1L x360%%.x30yr
SEES yee ae
ose = = 0-72pCi Laverage over 70 yrs
Esp Erz HQE se er 70.
oy „ One yt
PORT SOC
Risk =0:72pCi/ Lx Qomem/ ye, Leaneer death , rem cy 200017 = 0.2%
TSPCi/L “7 8000rem TO’ meer
7.47 20013 house, 0 2ach, oven+2bumers 6pm to Tpm, find CO pm and and 10pm, For
‘these circumstances, (7.62) is appropriate:
Sy ge
CW = Fle")
the source strength, Sis
67pm: oven + 2 bumers,1900mphr= 251540m/r = 5580mghr
solving for Cafter he:
(Up) = SOE NN) 16.8mg/
02 HEHEHE om”
Pe. 7.19
imers and watch CO coast down until 10pm, 3hrs later:
Now tum off the
CGhrs10pm) = C(0,7pm)e”*
1=0.39ach, V=27m3, after I-hr NO = 4.7ppm; find source strength, S:
7.48
first convert NO in ppm to mg/ using (15) and assuming T=259C,
gd ne IO ee à
24465 22465
a. Source strength, rearange (7.62)
wwe _ 039% x27 5.768
S= = he et = 188mgNO Ihr
= ( 7
b. I-hrafer tuning off the heater,
C= Ce = 4.7ppm xe V3 2ppmNO
€. in a house with 0.2 ach, 30013,
3.1m9/m'x 45 ppm
7.49. 100MW coal plant, 33.3% efficient, CF=0.70,
a. electricity generated per year,
100,000kW x 24hr/day x 365 daylyr x 0.70
Energy
b. heat input = 613x10/ kWh /yrourx Min „ 24080
LW, out * Wa
Shit and sei he allowances
Bu 96150, to
le "200
Saved by shutting down
Le , Lale, $2 mint
1883
ye
Pe. 720
613x10°KWh /yr
528x10"°Bww/yr
Now tum off the
CGhrs10pm) = C(0,7pm)e”*
1=0.39ach, V=27m3, after I-hr NO = 4.7ppm; find source strength, S:
7.48
first convert NO in ppm to mg/ using (15) and assuming T=259C,
gd ne IO ee à
24465 22465
a. Source strength, rearange (7.62)
wwe _ 039% x27 5.768
S= = he et = 188mgNO Ihr
= ( 7
b. I-hrafer tuning off the heater,
C= Ce = 4.7ppm xe V3 2ppmNO
€. in a house with 0.2 ach, 30013,
3.1m9/m'x 45 ppm
7.49. 100MW coal plant, 33.3% efficient, CF=0.70,
a. electricity generated per year,
100,000kW x 24hr/day x 365 daylyr x 0.70
Energy
b. heat input = 613x10/ kWh /yrourx Min „ 24080
LW, out * Wa
Shit and sei he allowances
Bu 96150, to
le "200
Saved by shutting down
Le , Lale, $2 mint
1883
ye
Pe. 720
613x10°KWh /yr
528x10"°Bww/yr
SOLUTIONS FOR CHAPTER 8
8.1. From (8.1),
(ah Taha ejac
OS IAS IL
sign means the sample corresponds to a warmer climate.
8.2 From the equation given for the ice core,
rec)
5 80(%,)+204=1.5x39+204= 1°C
notice, by the way, that since his sample is fr glacial ic, not ocean water or sedimnt,
the negative sign on 5"0(9/,) means colder temperatures
8.3 TheMateath!
1370W/m2
Essens” Encore
SIR =0T'A = TR)
To,
(5 sy. 1370W/ 0%
RSTO WI
8.4 The basicrelationshipis $= À, using data for Earth from Table 82 lets us find kı
eS 4? = 1370W Im? x (150x10°km x 10'm/ km)” = 3.089107
x 308310
& MERDE S= 35" Tao x 10 mn)
=916W/m°
b. The effective temperature (8.7) of Mercury would be:
= (za) [pompe
Rene eux ao
Ps. 81
8.1. From (8.1),
(ah Taha ejac
OS IAS IL
sign means the sample corresponds to a warmer climate.
8.2 From the equation given for the ice core,
rec)
5 80(%,)+204=1.5x39+204= 1°C
notice, by the way, that since his sample is fr glacial ic, not ocean water or sedimnt,
the negative sign on 5"0(9/,) means colder temperatures
8.3 TheMateath!
1370W/m2
Essens” Encore
SIR =0T'A = TR)
To,
(5 sy. 1370W/ 0%
RSTO WI
8.4 The basicrelationshipis $= À, using data for Earth from Table 82 lets us find kı
eS 4? = 1370W Im? x (150x10°km x 10'm/ km)” = 3.089107
x 308310
& MERDE S= 35" Tao x 10 mn)
=916W/m°
b. The effective temperature (8.7) of Mercury would be:
= (za) [pompe
Rene eux ao
Ps. 81
e. Peak wavelength
Aou =
Solar flux variation of £3.3%, gives a range of S
Smax = 1370 (140.033) = 14152 Wim?
13248 Wim?
Smin = 1370 (1 -0.033)
[su-ay Y passat cosy!
oro WIR] “2562 CITO
Uso
292K (aro
EST
4x5.67x10W/ m°K°
a difference of about 4°C,
Anuelear wor:
COMINO 342 w/m
renuscrep RADIATED TO SPACE, x
ES tt Î
vos
+) cowreenon,
pe EVAPORATION = 0
EARTH © RADIATED | ©
20 win?
a. Surface temperature,
OT = 240W Im!
T 255K (180)
[ 240W/ m? ]
ES
b. X, atmosphere to space: balance incoming from space =
going to space
342 =69+X X=273 Wim?
©. Y, absorbed by earth: incoming solar has to go somewhere,
Aou =
Solar flux variation of £3.3%, gives a range of S
Smax = 1370 (140.033) = 14152 Wim?
13248 Wim?
Smin = 1370 (1 -0.033)
[su-ay Y passat cosy!
oro WIR] “2562 CITO
Uso
292K (aro
EST
4x5.67x10W/ m°K°
a difference of about 4°C,
Anuelear wor:
COMINO 342 w/m
renuscrep RADIATED TO SPACE, x
ES tt Î
vos
+) cowreenon,
pe EVAPORATION = 0
EARTH © RADIATED | ©
20 win?
a. Surface temperature,
OT = 240W Im!
T 255K (180)
[ 240W/ m? ]
ES
b. X, atmosphere to space: balance incoming from space =
going to space
342 =69+X X=273 Wim?
©. Y, absorbed by earth: incoming solar has to go somewhere,
342 = 69+257+¥ Y= 16 Wim?
4. Z, radiation from atmosphere to surface: balance earth's surface radiation,
Y+Z=240=16+Z Z=224 Wim?
8.7. A2ayeratmosphere:
se so
un] wh
(Ait 1
DRE
350
y rw)
168 sol
a. atthe surface: 168+ Z = 24 + 78 +390 Z=324 Wim?
b. extraterrestrial: 342 = 107+ 40+W W= 195 Wim?
€: lomeratmosphee
Y +244 78 +350 + 195=2 x 324 Y=1 Wim?
4. incoming: 342=107+X+1+168 X=66 Wim?
+. temperatures T and Ta, from
a 195W/m° a ei
oT! = Wa 195 +(e) PAR (SO)
324Wim® a
pf eZ 2324 n(n 75K (2°C)
8.8 Moped
un,
W
2465kJ/kg x 10° kg/m? x 10'J/kJ
FEW Im? xSAXIO RE x
3600 124 À x365,
evaporation = som
averaged over de glo, with area 5.141014 m? ie very clos o anal
precipitation.
4. Z, radiation from atmosphere to surface: balance earth's surface radiation,
Y+Z=240=16+Z Z=224 Wim?
8.7. A2ayeratmosphere:
se so
un] wh
(Ait 1
DRE
350
y rw)
168 sol
a. atthe surface: 168+ Z = 24 + 78 +390 Z=324 Wim?
b. extraterrestrial: 342 = 107+ 40+W W= 195 Wim?
€: lomeratmosphee
Y +244 78 +350 + 195=2 x 324 Y=1 Wim?
4. incoming: 342=107+X+1+168 X=66 Wim?
+. temperatures T and Ta, from
a 195W/m° a ei
oT! = Wa 195 +(e) PAR (SO)
324Wim® a
pf eZ 2324 n(n 75K (2°C)
8.8 Moped
un,
W
2465kJ/kg x 10° kg/m? x 10'J/kJ
FEW Im? xSAXIO RE x
3600 124 À x365,
evaporation = som
averaged over de glo, with area 5.141014 m? ie very clos o anal
precipitation.
an
8.9. Greenhouse enhanced earth:
302 a
100. | |
ern
E
a. incoming energy: 342 = 100+ 67 + W W=175 Wim?
b. find Z from radiation to space:
342 = 100+30+Z Z=212 Wim?
©. to find X, need the energy radiated by a 291 K surface:
surface radiated = @T' = 56710" W/m" -K* x (291K) = 406.6W/ m"
376.6 Wim?
sothat, 406.6= X + 30
4. can find Y several ways: atthe surface, orin the atmosphere,
33.6 Wim?
W+Y=4066+24+78 = 175+Y
on, 67+24+78+X=Y+Z
67+24+78+3766 =Y+212 (Y=3336)
8.10 Solar constant § = 1370, increase by 0.1%, AS=0.001x 1370 = 137 Wim?
perunit area of earth,
change insadiationhittingearth _ 137W/m° x xR’
‘surface area of earth AR
ity of 0.55,
AF = 055x033 = 0.19°C
Ar
Climate sensi
ar
8.11 Changing the albedo from 031 10030 changes radiative forcing by
(031 - 030) x 342 Wim? = 3.42 Wim?
Total AF 742 Wim?
42 + A
am & 55 x 7,42 = 4.1 0C
AF
Pe. 84
8.9. Greenhouse enhanced earth:
302 a
100. | |
ern
E
a. incoming energy: 342 = 100+ 67 + W W=175 Wim?
b. find Z from radiation to space:
342 = 100+30+Z Z=212 Wim?
©. to find X, need the energy radiated by a 291 K surface:
surface radiated = @T' = 56710" W/m" -K* x (291K) = 406.6W/ m"
376.6 Wim?
sothat, 406.6= X + 30
4. can find Y several ways: atthe surface, orin the atmosphere,
33.6 Wim?
W+Y=4066+24+78 = 175+Y
on, 67+24+78+X=Y+Z
67+24+78+3766 =Y+212 (Y=3336)
8.10 Solar constant § = 1370, increase by 0.1%, AS=0.001x 1370 = 137 Wim?
perunit area of earth,
change insadiationhittingearth _ 137W/m° x xR’
‘surface area of earth AR
ity of 0.55,
AF = 055x033 = 0.19°C
Ar
Climate sensi
ar
8.11 Changing the albedo from 031 10030 changes radiative forcing by
(031 - 030) x 342 Wim? = 3.42 Wim?
Total AF 742 Wim?
42 + A
am & 55 x 7,42 = 4.1 0C
AF
Pe. 84
8.12 a. C3HF7 isan HFC (no Cl), 317 -90=227, HFC-227
b. CHH3FCI isan HCFC, 231-
=141, HCFC-14t
e. CoFaCla isa CFC, 204-90=114, CFC-114
4. CF3Br is a Halon, H-1301
8.13 a, HCFC-225, 225+ 90=315 (3C, 1H, SF), Ssites-(145)=2Cl, x. C3HFSCI2
b. HFC32, 32+90=122 (IC, 2H, 2F) 4 sites, O CL, +. CHP2
e. He1301, (IC, 3F, OCA, Br) 2 CE3Br
4. CFC-114, 114 + 90 = 204 (2C, OH, 4F), 6 sites -4= +. Fa
8.14 Radiative forcing forN20,
are e
! 0.14
ER TAE
ic machy #17 py ring wold bs
AF = (JE = JG) = 0.133087 - STI) =037 W/m?
133
8.15 Combined radiative forcing (ignoring the complication mentioned in Prob. 8.19):
12)
8,
Feu, =0.031( (CH, — CH), ) = 0.031(YE7TS — 700) - 0.463 Wim?
= 0.13 JNO - NO),
AFereur = O2ALCFC- 11)-(CFC-11),] =0.22(0.268 - 0) =0.059 W/m°
a. AF: =63 In 1.558 Wim?
= 0.13{ STI - 23) =0.140W /m°
AFcroux = 0.28{(CFC -12)-(CFC-12),]=0.28(0.503 ~0)= 0.141 W/m?
Combined forcing = 1.558 + 0.463 + 0.140 + 0.059 + 0.141 = 2.36 Wim?
Pg. 85
b. CHH3FCI isan HCFC, 231-
=141, HCFC-14t
e. CoFaCla isa CFC, 204-90=114, CFC-114
4. CF3Br is a Halon, H-1301
8.13 a, HCFC-225, 225+ 90=315 (3C, 1H, SF), Ssites-(145)=2Cl, x. C3HFSCI2
b. HFC32, 32+90=122 (IC, 2H, 2F) 4 sites, O CL, +. CHP2
e. He1301, (IC, 3F, OCA, Br) 2 CE3Br
4. CFC-114, 114 + 90 = 204 (2C, OH, 4F), 6 sites -4= +. Fa
8.14 Radiative forcing forN20,
are e
! 0.14
ER TAE
ic machy #17 py ring wold bs
AF = (JE = JG) = 0.133087 - STI) =037 W/m?
133
8.15 Combined radiative forcing (ignoring the complication mentioned in Prob. 8.19):
12)
8,
Feu, =0.031( (CH, — CH), ) = 0.031(YE7TS — 700) - 0.463 Wim?
= 0.13 JNO - NO),
AFereur = O2ALCFC- 11)-(CFC-11),] =0.22(0.268 - 0) =0.059 W/m°
a. AF: =63 In 1.558 Wim?
= 0.13{ STI - 23) =0.140W /m°
AFcroux = 0.28{(CFC -12)-(CFC-12),]=0.28(0.503 ~0)= 0.141 W/m?
Combined forcing = 1.558 + 0.463 + 0.140 + 0.059 + 0.141 = 2.36 Wim?
Pg. 85
b. From 1992 10 2100:
(0,)]
(Co.),
Ar, =001( CH - (CM
710)
Su) messi
0031816 - JITTA)=0.581W/00"
BF. = 0.131550 - YIN;O),}= 0.131417 - ST) =0.370W/ m?
Feu = 022{(CFC~ 11)- (CFC~11)] =0.22(0.040-0.268) = -0.050 Wim’
Feros = 0.28{(CFC -12)-(CFC-12),)=0.28(0.207 - 0.503) = -0.083 Wi
Combined forcing = 4.35 + 0.581 + 0370-0050 - 0.08 = 5.17 Win?
© aFe6ain(Z2) +0081(43618-/700)+0.154 07-75)
+ 0.220.040 + 0.28x0.207 = 7.53 Wim?
(altematively: AF = 2.36 + 5.17 =7.53 Wim?)
8.16 From Prob. 8.15,
ar 63 (DD) + 001493 RB) 0.139 a VS)
+ +022x0.040 + 0.280.207 = 7.53 Wim?
ATs = À AF = 0.57 0CKWIm?) x 7.53 Wim? = 43 0C
8.17 From Prob. 8.15
are 63 (20) + 0081(43876-700)+ 0.15 aT - 78)
+ 02240040 + 02810207 7.55 Win?
and from (8.32),
(CO: 2278422780 % >= 919ppm
Pe. 86
(0,)]
(Co.),
Ar, =001( CH - (CM
710)
Su) messi
0031816 - JITTA)=0.581W/00"
BF. = 0.131550 - YIN;O),}= 0.131417 - ST) =0.370W/ m?
Feu = 022{(CFC~ 11)- (CFC~11)] =0.22(0.040-0.268) = -0.050 Wim’
Feros = 0.28{(CFC -12)-(CFC-12),)=0.28(0.207 - 0.503) = -0.083 Wi
Combined forcing = 4.35 + 0.581 + 0370-0050 - 0.08 = 5.17 Win?
© aFe6ain(Z2) +0081(43618-/700)+0.154 07-75)
+ 0.220.040 + 0.28x0.207 = 7.53 Wim?
(altematively: AF = 2.36 + 5.17 =7.53 Wim?)
8.16 From Prob. 8.15,
ar 63 (DD) + 001493 RB) 0.139 a VS)
+ +022x0.040 + 0.280.207 = 7.53 Wim?
ATs = À AF = 0.57 0CKWIm?) x 7.53 Wim? = 43 0C
8.17 From Prob. 8.15
are 63 (20) + 0081(43876-700)+ 0.15 aT - 78)
+ 02240040 + 02810207 7.55 Win?
and from (8.32),
(CO: 2278422780 % >= 919ppm
Pe. 86
8.18 From (834)
AT, 200
-046'"C( Win)
rar
b. Forcing by CO2 alone (830): AI
«ratos arca)
'non-CO2 forcing AF = 4.85 - 3.70 = 1.15 Wim?
4. Equilibrium temperature change,
Au. AE [228
EIN
AF „3.70
€ Warming y oz: Si „AR
8.19 Overlapping absorption bands correction factor:
M=3616 Mas 1714. NT Nos3II
MSN.)
0471nf1+201x10 MN) +531x10 M MN) ]
LM No)=0471a[1+ 20110 17431) + 531% 10 1714171481) 7] 0.159
f(MN,)= 047Inf+2.01310%(3616311)""+53 1x10%36163616:311)""] =0.256
AF, =0030/ (CHE - J(CH),) ~ MN) + MN)
=0.056( 43616 - JTTA)-0.256-+0.159
(this resultisviruallythe same as obtained in Example 8.6)
.577 = 0.58
Pa. 87
AT, 200
-046'"C( Win)
rar
b. Forcing by CO2 alone (830): AI
«ratos arca)
'non-CO2 forcing AF = 4.85 - 3.70 = 1.15 Wim?
4. Equilibrium temperature change,
Au. AE [228
EIN
AF „3.70
€ Warming y oz: Si „AR
8.19 Overlapping absorption bands correction factor:
M=3616 Mas 1714. NT Nos3II
MSN.)
0471nf1+201x10 MN) +531x10 M MN) ]
LM No)=0471a[1+ 20110 17431) + 531% 10 1714171481) 7] 0.159
f(MN,)= 047Inf+2.01310%(3616311)""+53 1x10%36163616:311)""] =0.256
AF, =0030/ (CHE - J(CH),) ~ MN) + MN)
=0.056( 43616 - JTTA)-0.256-+0.159
(this resultisviruallythe same as obtained in Example 8.6)
.577 = 0.58
Pa. 87
racic
44700 0.65
17920 0.26
6.160 0.09
44,700 0.7
6720 0.11
680 012
50 CO2 470 1 4470 08
50 CHa 320 65 2,080 0.04
50 NO 2 170 37740 0.07
‘The actual AT realized is estimated to be about 0.69C, which is 75% of the equilibrium AT
ATrealized = 0.6°C = 0.75 ATequilibrium
1610.75 = 0.800
50, ATequilibriom
but AFacual = 0.57 x AFacnal = 08
that is, AFactual 140W/m°
“The directforcingis245 Wim? , so aerosols te are 245- 1.40 = 1.05 Wim?
8.23 Energy sources and carbon intensity:
Coal 25% @ 242 20M1
Gi 45% @ 197 30M
Gas 20% @13830M)
Other 10% @ 0
a. avg C intensity = 025x24.2 + 0.45% 19.7 + 0.20x13.8 + 0.10x0 = 17.68 gC/MI
b. Coal replaced by non-carbon emitting sources:
avg C intensity = 0.25x0 + 0.45x 19:7 + 0.20x13.8 + 0.10%
1.63 CMI
Pg. 88
44700 0.65
17920 0.26
6.160 0.09
44,700 0.7
6720 0.11
680 012
50 CO2 470 1 4470 08
50 CHa 320 65 2,080 0.04
50 NO 2 170 37740 0.07
‘The actual AT realized is estimated to be about 0.69C, which is 75% of the equilibrium AT
ATrealized = 0.6°C = 0.75 ATequilibrium
1610.75 = 0.800
50, ATequilibriom
but AFacual = 0.57 x AFacnal = 08
that is, AFactual 140W/m°
“The directforcingis245 Wim? , so aerosols te are 245- 1.40 = 1.05 Wim?
8.23 Energy sources and carbon intensity:
Coal 25% @ 242 20M1
Gi 45% @ 197 30M
Gas 20% @13830M)
Other 10% @ 0
a. avg C intensity = 025x24.2 + 0.45% 19.7 + 0.20x13.8 + 0.10x0 = 17.68 gC/MI
b. Coal replaced by non-carbon emitting sources:
avg C intensity = 0.25x0 + 0.45x 19:7 + 0.20x13.8 + 0.10%
1.63 CMI
Pg. 88
©. modeled as an exponential growth function,
0.0082 = ~0.42%! yr
8.24 Out of oil and gas, demand = 2x330El1yr, 28%coal, 60% syn gas/oil@44gC/MJ,
2. carbon emission rate:
avg carbon intensity = 0.28 x 258 + 0.60 x 44 + 0.12 x0 = 33.6 ¿C/MI
2x 33000, MI 3680 GC
ye NIT M Tome ROCA
b. growth from 6.0 GtC/yr to 22.2 G1C/yrin 100 yes,
sE (22)
Emission:
100 "(60
fraction, use (3.15):
nn. 89GICI yr,
(as
50x 1239=619 GC
totalemited=Q Sn 1) 123901C
amount remaining in atmosphere =
4d. amount in atmosphere in 100 yrs = 750 + 619 = 1369 GC
1369GC
2ARGIC/ ppm CO.
646ppm
+. equilibrium temperature increase, with AT94=3°C,
2.)
ina 3%.
8.25 Repeatof Prob. 8.24, but now conservation scenario:
a. carbon emission rate:
avg carbon intensity = 0.20 x 258 + 0.30 x 153 + 0.10 x20 = 11.75 gC/MI
Emissions FO, MI, 11.7530, GC
Fu MOT M ‘10%
=3:8801C/yr
Pg. 89
0.0082 = ~0.42%! yr
8.24 Out of oil and gas, demand = 2x330El1yr, 28%coal, 60% syn gas/oil@44gC/MJ,
2. carbon emission rate:
avg carbon intensity = 0.28 x 258 + 0.60 x 44 + 0.12 x0 = 33.6 ¿C/MI
2x 33000, MI 3680 GC
ye NIT M Tome ROCA
b. growth from 6.0 GtC/yr to 22.2 G1C/yrin 100 yes,
sE (22)
Emission:
100 "(60
fraction, use (3.15):
nn. 89GICI yr,
(as
50x 1239=619 GC
totalemited=Q Sn 1) 123901C
amount remaining in atmosphere =
4d. amount in atmosphere in 100 yrs = 750 + 619 = 1369 GC
1369GC
2ARGIC/ ppm CO.
646ppm
+. equilibrium temperature increase, with AT94=3°C,
2.)
ina 3%.
8.25 Repeatof Prob. 8.24, but now conservation scenario:
a. carbon emission rate:
avg carbon intensity = 0.20 x 258 + 0.30 x 153 + 0.10 x20 = 11.75 gC/MI
Emissions FO, MI, 11.7530, GC
Fu MOT M ‘10%
=3:8801C/yr
Pg. 89
b. growih from 6.0 GiClyr 10 3.88 GiClyr in 100 yrs,
L (E)--oon- 0.449%) yr
so
&. amount remaining with 50% aibome fraction, ue (2.15):
=p» OC
(en) LE
total emitted = Q m0
[a asie
amount remaining in atmosphere = 0.50 x 483 = 242.GtC
4d. amount in atmosphere in 100 yrs = 750 + 242= 992 GıC
MC
(OO Truc ppmcO,
= 468 ppm
+. equilibrium temperature increase, with AT24=3°C,
Sa
Ar:
in2 "|
8.26 Doitby scenario:
(A) P= 10403-20-0.7=-14%/yr
SOCIA (rem
Papes
ae le oad
TSOGIC + QxAF
(co,)= 2.12GtC/ ppmCO,
ER |.
60
Pe. 8.10
L (E)--oon- 0.449%) yr
so
&. amount remaining with 50% aibome fraction, ue (2.15):
=p» OC
(en) LE
total emitted = Q m0
[a asie
amount remaining in atmosphere = 0.50 x 483 = 242.GtC
4d. amount in atmosphere in 100 yrs = 750 + 242= 992 GıC
MC
(OO Truc ppmcO,
= 468 ppm
+. equilibrium temperature increase, with AT24=3°C,
Sa
Ar:
in2 "|
8.26 Doitby scenario:
(A) P= 10403-20-0.7=-14%/yr
SOCIA (rem
Papes
ae le oad
TSOGIC + QxAF
(co,)= 2.12GtC/ ppmCO,
ER |.
60
Pe. 8.10
(0)
154 15-02 +042 3.2%yr
EDGE yn:
002“
en) 1STAGC
TSOGIC +QrAF__ 750+ 1574x0.561C
(o.=z T2GICT ppmCO, "2.1201 /ppmco, ep
2 [75
Zu @)-2osc
0002
pa [o PEO A
NEO
© rata 10-10-02 12m
(er) = 658010
750+ 65830 561C
(Od 3 TaGici neo, ~2.12G1C/ppmco, ~ FP
|]
sel
10043
D ET
Tom
100s
2) cone gas —>[n-070 138086 1979
a 9.79073
heatpump _ 105MJ dei |
Dremtpump 100m re M ef cons Pm 4420.25 00cm
242090 dns
Power Pant F0 fom envio
resistance 100M) TGS LEM 282080 ac
242090 350 =
poner pant |
Pg, sul |
OS WB
154 15-02 +042 3.2%yr
EDGE yn:
002“
en) 1STAGC
TSOGIC +QrAF__ 750+ 1574x0.561C
(o.=z T2GICT ppmCO, "2.1201 /ppmco, ep
2 [75
Zu @)-2osc
0002
pa [o PEO A
NEO
© rata 10-10-02 12m
(er) = 658010
750+ 65830 561C
(Od 3 TaGici neo, ~2.12G1C/ppmco, ~ FP
|]
sel
10043
D ET
Tom
100s
2) cone gas —>[n-070 138086 1979
a 9.79073
heatpump _ 105MJ dei |
Dremtpump 100m re M ef cons Pm 4420.25 00cm
242090 dns
Power Pant F0 fom envio
resistance 100M) TGS LEM 282080 ac
242090 350 =
poner pant |
Pg, sul |
OS WB
Notice the tremendous range: 14,5 to 69.1 gC/MIJ, almost 5:1!
8.28 Propane-fired water heater:
a Gy, = 32128C/ mal
A sus
n =
100 1,2085
153690
©. savings vs 34,0 gC/MI] with an electric water heater:
propane _ 193
electric 340
157 so there is a43% savings
8.29 Using (3.18) to find © , and (3.20) to find tm „then plotting (3.17) gives.
2588C,_GIC_ 10%
wm oc ES
a Q, =200,000E1 x: =516061C
516061C
we 7 DGC er
ë BEC
BB an EE.
N » wa ae Ina
then put these into P = P, DE Ee ste)
3.573
Ca Es 2537 ET
Pr T 22) Et 38
Pro 367 367] EX
sigma. 33,57 50.55] 35.49]
[un 150.84) 112.77] 75.60]
[deta time o]
ar [2
7990) 500] 50] 50]
2000) 7.03) 8.05] 70.57
Pe. 8.12
8.28 Propane-fired water heater:
a Gy, = 32128C/ mal
A sus
n =
100 1,2085
153690
©. savings vs 34,0 gC/MI] with an electric water heater:
propane _ 193
electric 340
157 so there is a43% savings
8.29 Using (3.18) to find © , and (3.20) to find tm „then plotting (3.17) gives.
2588C,_GIC_ 10%
wm oc ES
a Q, =200,000E1 x: =516061C
516061C
we 7 DGC er
ë BEC
BB an EE.
N » wa ae Ina
then put these into P = P, DE Ee ste)
3.573
Ca Es 2537 ET
Pr T 22) Et 38
Pro 367 367] EX
sigma. 33,57 50.55] 35.49]
[un 150.84) 112.77] 75.60]
[deta time o]
ar [2
7990) 500] 50] 50]
2000) 7.03) 8.05] 70.57
Pe. 8.12
Prob. 8.29 Solution
"TE y T TT
2 1
ier dE]
3 TT
go = rt
Í
E T
=» i =e
ÿ » f
Ë "I i i
ET
1900 2000 2100 2200 2300 zum
Year
GC, ou
0 Q. =2 Be. = 51606
8.30 Q. =200,000E x AS
1ppmco,
100, 0:73 x LPPMCO _ 177
CO, = 516061 x 0:73 x PRESS - 1776 ppm
(C02 = 1776 + 280 = 2056 ppm
3 (2056
Te
orgy) ee
. 2xl2gClmol 10'KJ
sis CH: REECE 15.56eC/M
3x12gCimol 10%
BGs FERRI «AE = 16224C/MI
4e12gCimol 10*KI
ET My 1SSEC/M
A
off = 35%
Pg. 8.13
"TE y T TT
2 1
ier dE]
3 TT
go = rt
Í
E T
=» i =e
ÿ » f
Ë "I i i
ET
1900 2000 2100 2200 2300 zum
Year
GC, ou
0 Q. =2 Be. = 51606
8.30 Q. =200,000E x AS
1ppmco,
100, 0:73 x LPPMCO _ 177
CO, = 516061 x 0:73 x PRESS - 1776 ppm
(C02 = 1776 + 280 = 2056 ppm
3 (2056
Te
orgy) ee
. 2xl2gClmol 10'KJ
sis CH: REECE 15.56eC/M
3x12gCimol 10%
BGs FERRI «AE = 16224C/MI
4e12gCimol 10*KI
ET My 1SSEC/M
A
off = 35%
Pg. 8.13
assuming a 100% capacity factor (plant operates all of the time),
SOM. IMs, 36008 76 248 OMC 66 sohonneC yr
a hr yr MY “10 §C
E
$20
eee” 52 6milion /ye
tax = 1.08x10'omneC /yrx E
bo carbon sequesteri
OSO tonneC /yr 10
IOS 21, 600ucres
3000kgC/yr-acre “tonne
Area
©. biomass instead of tax,
$2.16x10'/yn
restry cost = =$100/yr peracre
Forestry cost = = $100/ yr
8.33. Leaky landfill, 10 tonnes CH per year:
a. 20:ÿr GWP for methane = 56,
10 tonnes CH,/yrx 56 = 560 tonnes CO (equivalent )
». buming the methane,
CH4 + 202 > C02 + 2120
COjemited = MOICO, , 12+ 2x16)tomneCO, Imol 1OtonneCH,
mol CH, * (12 +AxDienneCH,/mol ye
=27.5tonne CO, yr
©. equivalent CO savings = 560 - 27.5 = 532.5 tonne CO2
12 tonneC
asC: 532.StonneCO,/yrx EC ronneC /yr saved
©: $82.StonneCO,/ yrx LEE = 145 2tomneC /y
4. carbon tax saved = 1452 tonneCiyr x $20VtonneC =$2900// yr
«same thing, 532.Stonne COZ saved x 85.45tonneCO2 = $2900/yr
Pg. 8.14
SOM. IMs, 36008 76 248 OMC 66 sohonneC yr
a hr yr MY “10 §C
E
$20
eee” 52 6milion /ye
tax = 1.08x10'omneC /yrx E
bo carbon sequesteri
OSO tonneC /yr 10
IOS 21, 600ucres
3000kgC/yr-acre “tonne
Area
©. biomass instead of tax,
$2.16x10'/yn
restry cost = =$100/yr peracre
Forestry cost = = $100/ yr
8.33. Leaky landfill, 10 tonnes CH per year:
a. 20:ÿr GWP for methane = 56,
10 tonnes CH,/yrx 56 = 560 tonnes CO (equivalent )
». buming the methane,
CH4 + 202 > C02 + 2120
COjemited = MOICO, , 12+ 2x16)tomneCO, Imol 1OtonneCH,
mol CH, * (12 +AxDienneCH,/mol ye
=27.5tonne CO, yr
©. equivalent CO savings = 560 - 27.5 = 532.5 tonne CO2
12 tonneC
asC: 532.StonneCO,/yrx EC ronneC /yr saved
©: $82.StonneCO,/ yrx LEE = 145 2tomneC /y
4. carbon tax saved = 1452 tonneCiyr x $20VtonneC =$2900// yr
«same thing, 532.Stonne COZ saved x 85.45tonneCO2 = $2900/yr
Pg. 8.14
8.34 Gasoline C7H15, 6.15 Ibs/gal, fully combusted,
a gusting «SIS IDAS, (7512 =84) Ibs
pel sal (7x12. + 15x15 99) Ib gas ale
= 90.000miles 5:22 lbs C
Ci
al
17,394 15€ that willbe released
b. 4000 1b car, 10,000 milye,
= 12394 Ibs, 10. 000m
= 4348 Ibs Cyr
40.000 mi yr an
carbon _ _ 4348 lbs Clye
Vehicle wt. 40001
the car emits slightly more carbon per year than it weighs!
1.08
SAC sis
E = 50009 gl = 3.96 /
al 2000 Ibs C DEBE $3.98 Laat
€: carbon tax =
4. new car @40mpg. for 40.000 miles:
40.000 , 522 Ibsc 12,174 Ibs C saved
17,394 IbsC «
+. trading in the clunker forthe 40 mpg vehicle would save
sis
2000 1650
that is, those C offsets would save the utility $91, which they could spend to get the
clunker off the road.
tax savings = 12,174 IbsC x 91 per car
8.38 Electrics vs gasoline powered cars:
522 Ibs Cígal 10003
ige *22 Tbs
ba with the very efficient natural
pelt
ETS
€. with the typical old coal plant,
as-fired power pant,
electo emissions
Pg. 815
a gusting «SIS IDAS, (7512 =84) Ibs
pel sal (7x12. + 15x15 99) Ib gas ale
= 90.000miles 5:22 lbs C
Ci
al
17,394 15€ that willbe released
b. 4000 1b car, 10,000 milye,
= 12394 Ibs, 10. 000m
= 4348 Ibs Cyr
40.000 mi yr an
carbon _ _ 4348 lbs Clye
Vehicle wt. 40001
the car emits slightly more carbon per year than it weighs!
1.08
SAC sis
E = 50009 gl = 3.96 /
al 2000 Ibs C DEBE $3.98 Laat
€: carbon tax =
4. new car @40mpg. for 40.000 miles:
40.000 , 522 Ibsc 12,174 Ibs C saved
17,394 IbsC «
+. trading in the clunker forthe 40 mpg vehicle would save
sis
2000 1650
that is, those C offsets would save the utility $91, which they could spend to get the
clunker off the road.
tax savings = 12,174 IbsC x 91 per car
8.38 Electrics vs gasoline powered cars:
522 Ibs Cígal 10003
ige *22 Tbs
ba with the very efficient natural
pelt
ETS
€. with the typical old coal plant,
as-fired power pant,
electo emissions
Pg. 815
UWheatin s/s 36008
O 13 cooks pme
030kW electric out kWheatin hr Ee OnE
12.000) 248C MI kWh A
A Ga
kWh MI 10’) Smi dia
i ‘coal plant heat rate
| electric emissions =
So catho can be saved wit este cars when efficient natale power plans are
| assumed, bt othe (pin ol col pant vey hls ay srs ar
8.36 NO, +hv~NO+O
2,2 19x10 -m/ mol _ 1.19x107
vee ARR mol 306
8.37 O.+hv>0+0
E LAOXIO" KI: m/ moi _ 119x107
HT mol 295
= 240x10°m =240nm
Bg, 8.16
O 13 cooks pme
030kW electric out kWheatin hr Ee OnE
12.000) 248C MI kWh A
A Ga
kWh MI 10’) Smi dia
i ‘coal plant heat rate
| electric emissions =
So catho can be saved wit este cars when efficient natale power plans are
| assumed, bt othe (pin ol col pant vey hls ay srs ar
8.36 NO, +hv~NO+O
2,2 19x10 -m/ mol _ 1.19x107
vee ARR mol 306
8.37 O.+hv>0+0
E LAOXIO" KI: m/ moi _ 119x107
HT mol 295
= 240x10°m =240nm
Bg, 8.16
SOLUTIONS FOR CHAPTER 9
A 30ÿ63 packer ruck, 750 (yd, 100 between stops, Smph, 1 min to load 200 Ib
ni tr, nin
a pea 5280 5m hr y e
* Ib ‚stop
Rein Ge 1.227min (6 If
ob. me be
émet =30y6750-%, x A, ME a
wal ie yd’ 163 lbs 60min
customers _ 30yd" x 750 Ib/yd* x 4homes/stop.
truckload 200 Ib/stop ah truellcaa!
Bo euionen 21008, 588 cateo |
mme = Beten, 21, Bogen |
Roue ii
Br ES
war] [| smn
a. time not on route
Lo il atruck takes:
25y¢" truc
xin truck 0230
two loads per day takes:
[oa rst meen cH
b. Customers = 25yd truck x.
# customers =:
S40, Shes $60, Ihr), Stay, S2wks
e. labor =| $2,
hr
4 ydcurb
Yen tack
500 customers 2
‘ruckead
day hr “day J wk” yr
Pg. 9.1
min + 3x20min + 2x15min + 15min + Ami
4 ya'eurb customer stop Si
"curb * deustomer "stop Toad
br
Obes day
customer
Garda = 20eusiomen ond
loads Sdays
day “week
0 nd,
A 30ÿ63 packer ruck, 750 (yd, 100 between stops, Smph, 1 min to load 200 Ib
ni tr, nin
a pea 5280 5m hr y e
* Ib ‚stop
Rein Ge 1.227min (6 If
ob. me be
émet =30y6750-%, x A, ME a
wal ie yd’ 163 lbs 60min
customers _ 30yd" x 750 Ib/yd* x 4homes/stop.
truckload 200 Ib/stop ah truellcaa!
Bo euionen 21008, 588 cateo |
mme = Beten, 21, Bogen |
Roue ii
Br ES
war] [| smn
a. time not on route
Lo il atruck takes:
25y¢" truc
xin truck 0230
two loads per day takes:
[oa rst meen cH
b. Customers = 25yd truck x.
# customers =:
S40, Shes $60, Ihr), Stay, S2wks
e. labor =| $2,
hr
4 ydcurb
Yen tack
500 customers 2
‘ruckead
day hr “day J wk” yr
Pg. 9.1
min + 3x20min + 2x15min + 15min + Ami
4 ya'eurb customer stop Si
"curb * deustomer "stop Toad
br
Obes day
customer
Garda = 20eusiomen ond
loads Sdays
day “week
0 nd,
+ SU 05yd? =$97,500/ yr
(898500 + $97,509" 55996) yp
Customer cost = pees
9.3 To avoid overtime pay, working 8 hrs per day and needing 165min to make the runs
back and forth tothe disposal ite, breaks, ete (see Prob. 9.2),
collection time = 8hr x 60min/hr- 165min =315 min/day
15min _stop_, deustomers | Sday
day "Sm stop “week NS
customers.
notice the annual cost of service per customeris now
SAO/MrxBhrldxSd/wkxS2wE/yr+S97,500/4200=543/yr
uck can be used. Asin 93,
9.4 So, with E-hr days a smaller
collection time = Shr x 6Omin/hr- 16Smin = 315 min/day
with 2 wuckloads per day,
315min _stop_, customers
ene _ gapeustomers/ da
day 1.5min ‘stop js ee
customers
or 420 customers pertruckload. At 2loads per day and 5 days per week, that would
‘give 4200 customers once a week service. Truck size needed is therefore,
420customers 02yd'at cur. yd? fa ruck
truckload customer Ad ateurs 229
truck iz
w
520 Sm) Sy Su
aor [0 Sa), St, Tue, 55.206 y
@ =) TE
($83,200+ 33 Or, 659 69 yp
4200 customers
{compared with $39.26 per customer in Prob. 9.2)
Pg 92
(898500 + $97,509" 55996) yp
Customer cost = pees
9.3 To avoid overtime pay, working 8 hrs per day and needing 165min to make the runs
back and forth tothe disposal ite, breaks, ete (see Prob. 9.2),
collection time = 8hr x 60min/hr- 165min =315 min/day
15min _stop_, deustomers | Sday
day "Sm stop “week NS
customers.
notice the annual cost of service per customeris now
SAO/MrxBhrldxSd/wkxS2wE/yr+S97,500/4200=543/yr
uck can be used. Asin 93,
9.4 So, with E-hr days a smaller
collection time = Shr x 6Omin/hr- 16Smin = 315 min/day
with 2 wuckloads per day,
315min _stop_, customers
ene _ gapeustomers/ da
day 1.5min ‘stop js ee
customers
or 420 customers pertruckload. At 2loads per day and 5 days per week, that would
‘give 4200 customers once a week service. Truck size needed is therefore,
420customers 02yd'at cur. yd? fa ruck
truckload customer Ad ateurs 229
truck iz
w
520 Sm) Sy Su
aor [0 Sa), St, Tue, 55.206 y
@ =) TE
($83,200+ 33 Or, 659 69 yp
4200 customers
{compared with $39.26 per customer in Prob. 9.2)
Pg 92
9.5 Comparing two truck sizes,
a. customers foreach truck:
(A) 27m truck
4 micarb__ customer
27m truck
# customers =
(B) 15m3 truck:
A m'eurb customer
Tin wuck “0250 cu
15m? truck x = 240customers /load
240 customers, 3 loads
=3600customers (15m?)
a a 1600customers (15m?)
customers =
b. hours per day for the crew:
(A) 27m? truck
customers | O4min_ 2 loads
= load customer day Homie
he
(246min+60min)x = 6 43h! day
(8) 15m? truck:
customers, O4min „3 loads
240: min
Toad“ usiomer day
(288min +215min)x IT — 8 38hrs/ da
“ Sain ”
<. cost per customer
(A) 27m3 truck:
hr =
22, $120 0007 yr = $207,672 /yr (27m)
$207,672/ yr
320eustomers
‘cost per customer S4807/yr (27m3)
(8) 15m3 truck:
a. customers foreach truck:
(A) 27m truck
4 micarb__ customer
27m truck
# customers =
(B) 15m3 truck:
A m'eurb customer
Tin wuck “0250 cu
15m? truck x = 240customers /load
240 customers, 3 loads
=3600customers (15m?)
a a 1600customers (15m?)
customers =
b. hours per day for the crew:
(A) 27m? truck
customers | O4min_ 2 loads
= load customer day Homie
he
(246min+60min)x = 6 43h! day
(8) 15m? truck:
customers, O4min „3 loads
240: min
Toad“ usiomer day
(288min +215min)x IT — 8 38hrs/ da
“ Sain ”
<. cost per customer
(A) 27m3 truck:
hr =
22, $120 0007 yr = $207,672 /yr (27m)
$207,672/ yr
320eustomers
‘cost per customer S4807/yr (27m3)
(8) 15m3 truck:
$40 S38hr Sday S2Wk
he day wk yr
SIS7AS2/ yr
cost per customer = A = $43.65/ yr (15m)
Pereustomer = Seopcustomers "S65" 9" (15m?)
'$70,000/ yr = $157,152/ yr
‘The 15m3 truck results in a lower cost of customer ser
9.6 A $150,000 truck, 2gal/mi, $1.50/gal, 10,000milyr, $20k/yr maintenance:
a. amortized at 12%, Bss
cars [it =
[2282812 can
(1+0.12)'=1
amortization
150,000 x 0201/yr = $30.195iy¢
10,000mi 2gal_ $1.50
ym gal
30,195 + 30.000 + 20.000 (maint) = $80.195/ye
fuel =
30,0001 ye
total truck cos
$25 2pcople 4Ohr S2wk
D hors PRE 104,000 yr
0 costa ~S80:1954$108.000/ o tome
Totonnes day x 260day/yr
9.7 Reworking Examples9.1 -93, using:
a. run per day,
hr - (gar to route) {route to dump) - (time at dump) - (dump to gar) - (breaks)
time left to collect =8-0.4-0.4-0.2-025-
75 hs
which allows N stops per day (1 load per day),
575014 x 3600%/hr
‘Gosistop x 1 loadiday
-=345st0ps/ load
truck volume needed is,
y a 0251 Istopx 345 opel _ 9 64 9
3.5 m’curb min truck
with economies,
he day wk yr
SIS7AS2/ yr
cost per customer = A = $43.65/ yr (15m)
Pereustomer = Seopcustomers "S65" 9" (15m?)
'$70,000/ yr = $157,152/ yr
‘The 15m3 truck results in a lower cost of customer ser
9.6 A $150,000 truck, 2gal/mi, $1.50/gal, 10,000milyr, $20k/yr maintenance:
a. amortized at 12%, Bss
cars [it =
[2282812 can
(1+0.12)'=1
amortization
150,000 x 0201/yr = $30.195iy¢
10,000mi 2gal_ $1.50
ym gal
30,195 + 30.000 + 20.000 (maint) = $80.195/ye
fuel =
30,0001 ye
total truck cos
$25 2pcople 4Ohr S2wk
D hors PRE 104,000 yr
0 costa ~S80:1954$108.000/ o tome
Totonnes day x 260day/yr
9.7 Reworking Examples9.1 -93, using:
a. run per day,
hr - (gar to route) {route to dump) - (time at dump) - (dump to gar) - (breaks)
time left to collect =8-0.4-0.4-0.2-025-
75 hs
which allows N stops per day (1 load per day),
575014 x 3600%/hr
‘Gosistop x 1 loadiday
-=345st0ps/ load
truck volume needed is,
y a 0251 Istopx 345 opel _ 9 64 9
3.5 m’curb min truck
with economies,
labor = $62,400/yr as bef
fore,
truck = $10,000 + 4000x24.64 = $108,570/yr
serving 345 stopslond x 1 ondday x 5 days = 1725 customers
RO SIT 5915 cto
refuse 25 se Mano astomenn Que 2691 tonne! yr
costing ITA Sse (lager wih Tai)
Serums per day
hr - (gar to route) «(route to dumpx3)- (me at dumpx3) - (dump to gar) - (breaks)
time left co collect = 8 - 0.4 - 0.4x5 -0.2x3 - 0.25
which allows N stops per day (1 load per day),
3.75hr/d x 36008 hr
‘GOsistop x 3 load/day
truck volume needed is,
y. 225m" stop x 75 st
3.5 m’eurb/m’in truck
with economies,
labor = $62,400/yr as bef
truck
= T5sops/load
sopaload _
536m"
fore,
10,000 + 4000x5.36= $31,428%yr
serving 75 stops/load x 3 loadiday x 5 days/w
75s
125 customers
$31,428 +62, 400/ yr _ $93,628
res Trader” 83-40/yr per customer
m wk 12048 tonne
a sn esha OOo HS Seale
Ve yg 20 12 Shoes x TET = 1755 tonne /
$93,828) ye
= $53.46/ tonne es with Table 9
7755 tonaeiye = 5-46! (all agrees with Table 9.9)
Pe.95
fore,
truck = $10,000 + 4000x24.64 = $108,570/yr
serving 345 stopslond x 1 ondday x 5 days = 1725 customers
RO SIT 5915 cto
refuse 25 se Mano astomenn Que 2691 tonne! yr
costing ITA Sse (lager wih Tai)
Serums per day
hr - (gar to route) «(route to dumpx3)- (me at dumpx3) - (dump to gar) - (breaks)
time left co collect = 8 - 0.4 - 0.4x5 -0.2x3 - 0.25
which allows N stops per day (1 load per day),
3.75hr/d x 36008 hr
‘GOsistop x 3 load/day
truck volume needed is,
y. 225m" stop x 75 st
3.5 m’eurb/m’in truck
with economies,
labor = $62,400/yr as bef
truck
= T5sops/load
sopaload _
536m"
fore,
10,000 + 4000x5.36= $31,428%yr
serving 75 stops/load x 3 loadiday x 5 days/w
75s
125 customers
$31,428 +62, 400/ yr _ $93,628
res Trader” 83-40/yr per customer
m wk 12048 tonne
a sn esha OOo HS Seale
Ve yg 20 12 Shoes x TET = 1755 tonne /
$93,828) ye
= $53.46/ tonne es with Table 9
7755 tonaeiye = 5-46! (all agrees with Table 9.9)
Pe.95
9.8 200 tonnes/d, Sd/wk, S3million, $100,000/3r, trucks $120,000, 20 tonneltrip, S8Ok/ÿr,
A triplday, 5 diwk, 10%, lyr amonization,
Staion costs: CRF(LOyr.10%)=| ia | 1627513
wy
‘SSmillion x 0.16275/yr + $100,000!yr = $588,236/yr
en]
(EU
day gtk
handle: 200 22 52K = 52, 000tomnes/yr
to hand Paket y
55882361 yr
whichis 358823619" 511 51/tonne
aoa 52,000tonnes/ yr pra
Truck costs:
depreciation= $120,000xCRF(10yr, 10%) = $120,000%0.16275/ y
driver + maintenance + depreciation = $80,000 + 19,530 = $99,530/yr
tohaut 20 SOMES ED 55 SOY x59 ™K 20 800tonnes/ truck = yr
ip aay rm
whichis BRON = 54:79 /tonne
20.800t0nne Iyr
fora total of $479+$113]
$16.10 per tonne
9.9 a. Costof direct haul to the disposal site,
540 +301 =40+30x 1.5 = $85itonne
». transfer station 0:3 hr from collection route,
$40 + 30 11 = 40 + 30x03 = $49 tonne to get tothe transfer station
forthe transfer station,
$10+ 1012
10+ 10 (1.5.hr- 03 he) = $22/ tonne
total cost = $49 + 2
7 tonne
€: minimum distance from the transfer station tothe disposal site,
directhaulS = Sto transfer station + Sor trnster station
$40 + S30/hrL.5hr = (540 + 30 11) + $10 10(15-11)
85-40 + 10 +15+30u- 1011 =654200
Pg. 96
19.530/yr
A triplday, 5 diwk, 10%, lyr amonization,
Staion costs: CRF(LOyr.10%)=| ia | 1627513
wy
‘SSmillion x 0.16275/yr + $100,000!yr = $588,236/yr
en]
(EU
day gtk
handle: 200 22 52K = 52, 000tomnes/yr
to hand Paket y
55882361 yr
whichis 358823619" 511 51/tonne
aoa 52,000tonnes/ yr pra
Truck costs:
depreciation= $120,000xCRF(10yr, 10%) = $120,000%0.16275/ y
driver + maintenance + depreciation = $80,000 + 19,530 = $99,530/yr
tohaut 20 SOMES ED 55 SOY x59 ™K 20 800tonnes/ truck = yr
ip aay rm
whichis BRON = 54:79 /tonne
20.800t0nne Iyr
fora total of $479+$113]
$16.10 per tonne
9.9 a. Costof direct haul to the disposal site,
540 +301 =40+30x 1.5 = $85itonne
». transfer station 0:3 hr from collection route,
$40 + 30 11 = 40 + 30x03 = $49 tonne to get tothe transfer station
forthe transfer station,
$10+ 1012
10+ 10 (1.5.hr- 03 he) = $22/ tonne
total cost = $49 + 2
7 tonne
€: minimum distance from the transfer station tothe disposal site,
directhaulS = Sto transfer station + Sor trnster station
$40 + S30/hrL.5hr = (540 + 30 11) + $10 10(15-11)
85-40 + 10 +15+30u- 1011 =654200
Pg. 96
19.530/yr
9.10
9.11
u=ihr 15-1
(hat is, the transfer station must beat least 0.5 hrs away from the disposal site
At 50% recycling rate, a 16 can has 8 o new aluminum and 8 gfrecyeledaluminum.
Using data from tables. o
i
2107
10
x5iso Lar
RSR Na: Se
new aluminum = 8g x 235,000
880k
recycled aluminum
total energy require
1880 + 41.2= 1921 Kllcan forthe aluminum
From Table 9.13,
energylean = 1921 KI + 421 + 4 + 955 + 9greoyoledx PL
108
perliter, — energ
Heavier cans from yesteryear, 0.0205 kg/can, and at 25% recycling rate, says
new aluminum per can is 0.75 x 20.5g/can
and the recycled amount is 20.5-15375=5.125 g,
153758.
newaluminum = 15375g x 235000 4x4. =3613k1
2107
ceyletteminon = 5.2553 5150 Lx ass
total energy for aluminum =3613 + 26 = 3639 klican
From Example 9.5, primary energy for aluminum today is about 1443 kl/ean
1443KJ/ con
SBT = 0396 so the savings is a bit over 60% !
Pa. 9.7
9.11
u=ihr 15-1
(hat is, the transfer station must beat least 0.5 hrs away from the disposal site
At 50% recycling rate, a 16 can has 8 o new aluminum and 8 gfrecyeledaluminum.
Using data from tables. o
i
2107
10
x5iso Lar
RSR Na: Se
new aluminum = 8g x 235,000
880k
recycled aluminum
total energy require
1880 + 41.2= 1921 Kllcan forthe aluminum
From Table 9.13,
energylean = 1921 KI + 421 + 4 + 955 + 9greoyoledx PL
108
perliter, — energ
Heavier cans from yesteryear, 0.0205 kg/can, and at 25% recycling rate, says
new aluminum per can is 0.75 x 20.5g/can
and the recycled amount is 20.5-15375=5.125 g,
153758.
newaluminum = 15375g x 235000 4x4. =3613k1
2107
ceyletteminon = 5.2553 5150 Lx ass
total energy for aluminum =3613 + 26 = 3639 klican
From Example 9.5, primary energy for aluminum today is about 1443 kl/ean
1443KJ/ con
SBT = 0396 so the savings is a bit over 60% !
Pa. 9.7
9.12 US. using 1:6x106 tonnes/yr Al, 63% recycling rate,
a Pinan rain,
sev min = 037 16x10 tonne 100 KE x 25000 ao
cdi = 063 160 ex 10 0 E 200%
“Total primary energy for aluminum =(139.1 + 52)x1012 = 1443 x 101219
b. With no recycling,
all new aluminum = 1.6x10'tonne x 1000:
kg
Tome
©. COZ emissions (Table 9.12)
1) With recyeling:
new aluminur
75x10" tonneCO,
recycled Al = 0.63 x 1.6x10'tonneAl x 0.48 048x10%tonneCO,
Total = (7.75 + 0.48)x106 = 8.23x106 tonne CO2 Iyr
2) Without recycling:
tonne,
x10 tonel x 13.
{6x10 tonne x 13.1 SEE.
= 20.96810 tonneCO ¿/yr
CO reduction with recycling = (20.96 - 8.23)x106 tonne = 12.7 x106 tonne COp/yr
9.13, 3 million tonnes/yrof Al, 35% recovery rate,
a. COzemissions:
tonneCO,
new Al = 0.65 x 3.0x10°tonneAl x 13.120
Al = 0.65 x 3.0x10'tomneAl x 13.
F2 = 25.5410" onneCO,/ yr
recycled AL= 0.35 x 3.0x10'tonneAl x 0.48 = 0.50x10°tonneC
tonneCO,
tonel
Total = (25.54 + 0.50)x106 = 26 x106 tonne COr/yr
Pg.98
a Pinan rain,
sev min = 037 16x10 tonne 100 KE x 25000 ao
cdi = 063 160 ex 10 0 E 200%
“Total primary energy for aluminum =(139.1 + 52)x1012 = 1443 x 101219
b. With no recycling,
all new aluminum = 1.6x10'tonne x 1000:
kg
Tome
©. COZ emissions (Table 9.12)
1) With recyeling:
new aluminur
75x10" tonneCO,
recycled Al = 0.63 x 1.6x10'tonneAl x 0.48 048x10%tonneCO,
Total = (7.75 + 0.48)x106 = 8.23x106 tonne CO2 Iyr
2) Without recycling:
tonne,
x10 tonel x 13.
{6x10 tonne x 13.1 SEE.
= 20.96810 tonneCO ¿/yr
CO reduction with recycling = (20.96 - 8.23)x106 tonne = 12.7 x106 tonne COp/yr
9.13, 3 million tonnes/yrof Al, 35% recovery rate,
a. COzemissions:
tonneCO,
new Al = 0.65 x 3.0x10°tonneAl x 13.120
Al = 0.65 x 3.0x10'tomneAl x 13.
F2 = 25.5410" onneCO,/ yr
recycled AL= 0.35 x 3.0x10'tonneAl x 0.48 = 0.50x10°tonneC
tonneCO,
tonel
Total = (25.54 + 0.50)x106 = 26 x106 tonne COr/yr
Pg.98
b. Primary energy forthe aluminum:
new Al = 0.65 x 3.0x10'tonneAlx RC mann
rer A 035 x 3,Ou'tonmeA 5150 231000 KE = San ve
Total energy = (458 + 54)x1012 Kulyr= 465 x 1012 kr
9.14 Newsprint, 597% moisture, HHV=I8540KJ/kg, 6.1% H. Starting with 1 kg of as
received waste,
energy vaporizing moisture = 0.0597kgH.0 x 24404 = 145.64
kg as received: dry weight = 1- 0.0597 = 0.9403 kg
Hydrogen inthe dry waste = 0.061 x 0.9403 = 0.0574 kg
energy as H becomes H20 = 0.0574kgH x 20D 9440 - 1259.64
kgH kg
total energy lost in water vapor = 145.6 + 1259.6 = 1405kI/kg
LHV = HHV - 1405= 18,540 1 405 = 17,135 kg
9.18. Corugated boxes, 52% moisture, HHV=16380KI/kg, 57H in died materi
Using (28) Q =2440(W+9H)
W=0052 kglO/kg waste, — H =(1 - 0.052) x 0.057 =0.054 kgirkgwaste
KgH.O _ 1555 kt
gis 0.052 +9 005) pt
LHV = HHV- Qh = 16380 - 1313 = 15,067 kJ/kg
Q, = 2440:
kan. SAT ;
no he bot
o a“ kgH KgPET 10%, kg look tele
SHRED /bote xy
A ET 1) nn
a (os Een
Pg. 99
new Al = 0.65 x 3.0x10'tonneAlx RC mann
rer A 035 x 3,Ou'tonmeA 5150 231000 KE = San ve
Total energy = (458 + 54)x1012 Kulyr= 465 x 1012 kr
9.14 Newsprint, 597% moisture, HHV=I8540KJ/kg, 6.1% H. Starting with 1 kg of as
received waste,
energy vaporizing moisture = 0.0597kgH.0 x 24404 = 145.64
kg as received: dry weight = 1- 0.0597 = 0.9403 kg
Hydrogen inthe dry waste = 0.061 x 0.9403 = 0.0574 kg
energy as H becomes H20 = 0.0574kgH x 20D 9440 - 1259.64
kgH kg
total energy lost in water vapor = 145.6 + 1259.6 = 1405kI/kg
LHV = HHV - 1405= 18,540 1 405 = 17,135 kg
9.18. Corugated boxes, 52% moisture, HHV=16380KI/kg, 57H in died materi
Using (28) Q =2440(W+9H)
W=0052 kglO/kg waste, — H =(1 - 0.052) x 0.057 =0.054 kgirkgwaste
KgH.O _ 1555 kt
gis 0.052 +9 005) pt
LHV = HHV- Qh = 16380 - 1313 = 15,067 kJ/kg
Q, = 2440:
kan. SAT ;
no he bot
o a“ kgH KgPET 10%, kg look tele
SHRED /bote xy
A ET 1) nn
a (os Een
Pg. 99
9.17. Energy estimates based on HHV = 339(C) + 1440(H) - 139(0) + 105(5)
a. comugated bores: based on dry weight,
HHV(dry) = 339x373 + 1440%5.70 - 13954493 + 105x021 = 16,809 kg
“as received” there are (10,052) = 0.948 dry material kg of “as received”
HHV as received = 0.948 kg( ry) x 16.809 KTkg(ary) = 15935 Kg
6. junk mat
HHV(dry) = 3393787 + 14401541 - 139x4274 + 105x0.09 = 14,697 Ka
HHVGas received) = (1 - 0.0456)x14,697 = 14,027 kik,
«mined garbage:
HHV(ry) =339x44.99 + 1440x6.43 - 139x2876 + 105x0.52= 20.568 Kg
HHV(as received) = (1 - 0.72):20,568 = 5759 kg
de lawn grass:
HHV(dey) = 339x6.18 + 14404596 - 13913643 + 105x0.42 = 19,218 Akg
HHV(as received) = (1 -0.7524)x19.218 = 4758 kg
+. demolition softwood
HHV(dey) = 339x51.0 + 1440x62 - 139x41.8 + 105x0.1 = 20,417 kI/kg
HHV(as received) = (1 - 0.077)x20,417 = 18,845 kI/kg
E tires
HHV(dry) =339x79.1 + 1440x68 - 139x59 + 105x1.5 = 35,944 KI/kg
HHV(as received) = (1 - 0.0102)35,944 = 35,578 kI/kg
8. polystyrene:
HIV (dry) = 339187.10 + 1440x8.45 - 139:3.96 + 105x0.02 = 41,147 Klikg
HHV(as received) = (1 - 0.002)x41.147 = 41,064 ki/kg
Pa. 9.10
a. comugated bores: based on dry weight,
HHV(dry) = 339x373 + 1440%5.70 - 13954493 + 105x021 = 16,809 kg
“as received” there are (10,052) = 0.948 dry material kg of “as received”
HHV as received = 0.948 kg( ry) x 16.809 KTkg(ary) = 15935 Kg
6. junk mat
HHV(dry) = 3393787 + 14401541 - 139x4274 + 105x0.09 = 14,697 Ka
HHVGas received) = (1 - 0.0456)x14,697 = 14,027 kik,
«mined garbage:
HHV(ry) =339x44.99 + 1440x6.43 - 139x2876 + 105x0.52= 20.568 Kg
HHV(as received) = (1 - 0.72):20,568 = 5759 kg
de lawn grass:
HHV(dey) = 339x6.18 + 14404596 - 13913643 + 105x0.42 = 19,218 Akg
HHV(as received) = (1 -0.7524)x19.218 = 4758 kg
+. demolition softwood
HHV(dey) = 339x51.0 + 1440x62 - 139x41.8 + 105x0.1 = 20,417 kI/kg
HHV(as received) = (1 - 0.077)x20,417 = 18,845 kI/kg
E tires
HHV(dry) =339x79.1 + 1440x68 - 139x59 + 105x1.5 = 35,944 KI/kg
HHV(as received) = (1 - 0.0102)35,944 = 35,578 kI/kg
8. polystyrene:
HIV (dry) = 339187.10 + 1440x8.45 - 139:3.96 + 105x0.02 = 41,147 Klikg
HHV(as received) = (1 - 0.002)x41.147 = 41,064 ki/kg
Pa. 9.10
9.18. Draw the chemical structures:
a. 12347,8-hexachlorodiben2o-p-dioxin
a
a a
ES © \ a
a
b. 7,8-heptochlorodibenzo-p-dioxin
a
a à a
ci C à a
a a
+. octachlorodibenzo-p-dioxin
a a
als a
of G . a
a a
4. 2.3.4,78-pemtachlorodibenzofuran
a o a
po A,
a
©. 123,678-hexachlorodibenzofuran
a
a a
e > > a
al
Pg 9.11
a. 12347,8-hexachlorodiben2o-p-dioxin
a
a a
ES © \ a
a
b. 7,8-heptochlorodibenzo-p-dioxin
a
a à a
ci C à a
a a
+. octachlorodibenzo-p-dioxin
a a
als a
of G . a
a a
4. 2.3.4,78-pemtachlorodibenzofuran
a o a
po A,
a
©. 123,678-hexachlorodibenzofuran
a
a a
e > > a
al
Pg 9.11
9.19 US. 129million tons, 800 Ib/yd3, ell 10, 1 lifuyr, 80% is MSW, 1000 people:
yeton „200010 yd 271”
Y, ae
TS
29x10"
THRO? Iyr
@80% per cel,
Vine = STORIE oO TE lye
080
10910" yr acre
TTT Ta? = 24987» 25 000a6re8/ ye
Population of the U.S. is around 260 million, so the area per 1000 people is
24,987actes/ ye
= 0.1 ser! 1000 people «yr
.000 thousand people Oe 1000 x
Aa rer OP gg,
9.20 50,000 people, 40,000 tons/yr, 22% recovery, 1000 Ib/yd3, 10ft lift, S0%MSW:
Van 0 00 2-23.22!
fe land ,
Toa MSW” In’ ye
2.1UXI0'M / y
ORNE
Aus S210 ye x
ABS /yr
b. to complete the landfill will take:
40 Fe x2lis
time remaining = =16.5yr5
= DETTES
9.21 By increasing its recovery rate to 40 percent,
ton 20001 _yd 27, landfill
Vin = 40,0002 x(1 0,40) PUR, = 1.62x10°R"/ ye
ton “100016 ya “Ono MSW
Ag RUC nn xara ye
=21.5yrs (added 5 more years vs P9.20)
Pg, 9.12
yeton „200010 yd 271”
Y, ae
TS
29x10"
THRO? Iyr
@80% per cel,
Vine = STORIE oO TE lye
080
10910" yr acre
TTT Ta? = 24987» 25 000a6re8/ ye
Population of the U.S. is around 260 million, so the area per 1000 people is
24,987actes/ ye
= 0.1 ser! 1000 people «yr
.000 thousand people Oe 1000 x
Aa rer OP gg,
9.20 50,000 people, 40,000 tons/yr, 22% recovery, 1000 Ib/yd3, 10ft lift, S0%MSW:
Van 0 00 2-23.22!
fe land ,
Toa MSW” In’ ye
2.1UXI0'M / y
ORNE
Aus S210 ye x
ABS /yr
b. to complete the landfill will take:
40 Fe x2lis
time remaining = =16.5yr5
= DETTES
9.21 By increasing its recovery rate to 40 percent,
ton 20001 _yd 27, landfill
Vin = 40,0002 x(1 0,40) PUR, = 1.62x10°R"/ ye
ton “100016 ya “Ono MSW
Ag RUC nn xara ye
=21.5yrs (added 5 more years vs P9.20)
Pg, 9.12
9.22 Lawn trimmings, 620g moisture, 330 g decomposible organics represented by
C12.76H21.2809,26N0.54
1 mol of trimmings = 12x12.76 + 1x21.28 + 16x9.26 + 14x0.54 = 3302 g/mol
that i
1 kg of as received trimmings has 330g of decomposible organies (1 mole)
using (9.9) gives
Ci2.16H21.2809.26N0.54 + n H20 => m CHa +5CO2 + d NH3
where m = (4x12.76 + 21.28 - 249.26 -3x0.54V8 = 6.5225
so, 6.5225 moles of methane are produced per mole of dry trimmings, which is the same as
6.5225 moles of methane produced per kg of "as received” trimmings
0.0224m'CH, _ 6.5225moiCH,
molCH, kg as
2 Vo .146m'CH, /kg
5225 molCH, | 890%
b. CH, eneray= x 804 os
Ham: kg "asreceived" mol HRS
9.231 kg of food wastes has 720g of water and 280 g of dry CaHpOcNd
a € 45%
H 64%
O 28.8% 0.288x280 = 80.648
N 33% 0.033x280= 9248
Total =233.8g/mol
so a= 126/12
H: 1 g/mol xb mol
O: 16g/mot x ¢ mol
N: 14g/mol xd mol
“The chemical formula for dry food wastes: C10.5H17.9205.04N0.66
b. chemical reaction:
C10.5H17.9205,04N0.66 + n H20 > m CH4 + s CO2 + d NH3
where n = (410.5 -17.92 -2x5.04 +3x0,66)/4 = 3.995
m = (4x10.5 + 17.92 - 2x 5.04 - 3x0.66/8 = 5.9825
(4x10.5- 17.92 + 2x5.04 + 3x0.66)8 = 4.5175
6=0.66
Pg. 9.13
C12.76H21.2809,26N0.54
1 mol of trimmings = 12x12.76 + 1x21.28 + 16x9.26 + 14x0.54 = 3302 g/mol
that i
1 kg of as received trimmings has 330g of decomposible organies (1 mole)
using (9.9) gives
Ci2.16H21.2809.26N0.54 + n H20 => m CHa +5CO2 + d NH3
where m = (4x12.76 + 21.28 - 249.26 -3x0.54V8 = 6.5225
so, 6.5225 moles of methane are produced per mole of dry trimmings, which is the same as
6.5225 moles of methane produced per kg of "as received” trimmings
0.0224m'CH, _ 6.5225moiCH,
molCH, kg as
2 Vo .146m'CH, /kg
5225 molCH, | 890%
b. CH, eneray= x 804 os
Ham: kg "asreceived" mol HRS
9.231 kg of food wastes has 720g of water and 280 g of dry CaHpOcNd
a € 45%
H 64%
O 28.8% 0.288x280 = 80.648
N 33% 0.033x280= 9248
Total =233.8g/mol
so a= 126/12
H: 1 g/mol xb mol
O: 16g/mot x ¢ mol
N: 14g/mol xd mol
“The chemical formula for dry food wastes: C10.5H17.9205.04N0.66
b. chemical reaction:
C10.5H17.9205,04N0.66 + n H20 > m CH4 + s CO2 + d NH3
where n = (410.5 -17.92 -2x5.04 +3x0,66)/4 = 3.995
m = (4x10.5 + 17.92 - 2x 5.04 - 3x0.66/8 = 5.9825
(4x10.5- 17.92 + 2x5.04 + 3x0.66)8 = 4.5175
6=0.66
Pg. 9.13
©. CH,
C10.5H17.9205.04N0.66 + 3.995 H20 > 5.9825 CH4 + 4.5175 CO + 0.66 NH3
59825molesCH,
(59825+4,5175+0,66)males gas
3.6%
0.0224m'CH, 5982SmoICH,
34m
moICH, "kg "asreceived” o CH. ke
4. Ven
59825 molCH, | 890i)
kg "as received
as received”
mal
9.14
C10.5H17.9205.04N0.66 + 3.995 H20 > 5.9825 CH4 + 4.5175 CO + 0.66 NH3
59825molesCH,
(59825+4,5175+0,66)males gas
3.6%
0.0224m'CH, 5982SmoICH,
34m
moICH, "kg "asreceived” o CH. ke
4. Ven
59825 molCH, | 890i)
kg "as received
as received”
mal
9.14
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